The Project Gutenberg EBook of A Primer of Quaternions, by Arthur S. Hathaway Copyright laws are changing all over the world. Be sure to check the copyright laws for your country before downloading or redistributing this or any other Project Gutenberg eBook. This header should be the first thing seen when viewing this Project Gutenberg file. Please do not remove it. Do not change or edit the header without written permission. Please read the "legal small print," and other information about the eBook and Project Gutenberg at the bottom of this file. Included is important information about your specific rights and restrictions in how the file may be used. You can also find out about how to make a donation to Project Gutenberg, and how to get involved.
**Welcome To The World of Free Plain Vanilla Electronic Texts** **eBooks Readable By Both Humans and By Computers, Since 1971** *****These eBooks Were Prepared By Thousands of Volunteers!*****
Title: A Primer of Quaternions Author: Arthur S. Hathaway Release Date: February, 2006 [EBook #9934] [Yes, we are more than one year ahead of schedule] [This file was first posted on November 1, 2003] Edition: 10 Language: English Character set encoding: TeX *** START OF THE PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS *** E-text prepared by Cornell University, Joshua Hutchinson, John Hagerson, and the Online Distributed Proofreading Team.
ii
A PRIMER OF QUATERNIONS BY
ARTHUR S. HATHAWAY PROFESSOR OF MATHEMATICS IN THE ROSE POLYTECHNIC INSTITUTE, TERRE HAUTE, IND.
1896
iii
Preface The Theory of Quaternions is due to Sir William Rowan Hamilton, Royal Astronomer of Ireland, who presented his first paper on the subject to the Royal Irish Academy in 1843. His Lectures on Quaternions were published in 1853, and his Elements, in 1866, shortly after his death. The Elements of Quaternions by Tait is the accepted text-book for advanced students. The following development of the theory is prepared for average students with a thorough knowledge of the elements of algebra and geometry, and is believed to be a simple and elementary treatment founded directly upon the fundamental ideas of the subject. This theory is applied in the more advanced examples to develop the principal formulas of trigonometry and solid analytical geometry, and the general properties and classification of surfaces of second order. In the endeavour to bring out the number idea of Quaternions, and at the same time retain the established nomenclature of the analysis, I have found it necessary to abandon the term “vector ” for a directed length. I adopt instead Clifford’s suggestive name of “step,” leaving to “vector ” the sole meaning of “right quaternion.” This brings out clearly the relations of this number and line, and emphasizes the fact that Quaternions is a natural extension of our fundamental ideas of number, that is subject to ordinary principles of geometric representation, rather than an artificial species of geometrical algebra. The physical conceptions and the breadth of idea that the subject of Quaternions will develop are, of themselves, sufficient reward for its study. At the same time, the power, directness, and simplicity of its analysis cannot fail to prove useful in all physical and geometrical investigations, to those who have thoroughly grasped its principles. On account of the universal use of analytical geometry, many examples have been given to show that Quaternions in its semi-cartesian form is a direct development of that subject. In fact, the present work is the outcome of lectures that I have given to my classes for a number of years past as the equivalent of the usual instruction in the analytical geometry of space. The main features of this primer were therefore developed in the laboratory of the class-room, and I desire to express my thanks to the members of my classes, wherever they may be, for the interest that they have shown, and the readiness with which they have expressed their difficulties, as it has been a constant source of encouragement and assistance in my work. I am also otherwise indebted to two of my students,—to Mr. H. B. Stilz for the accurate construction of the diagrams, and to Mr. G. Willius for the plan (upon the cover) of the plagiograph or mechanical quaternion multiplier which was made by him while taking this subject. The theory of this instrument is contained in the step proportions that are given with the diagram.1 ARTHUR S. HATHAWAY. 1 See
Example 19, Chapter I.
Contents 1 Steps Definitions and Theorems Centre of Gravity . . . . . Curve Tracing, Tangents . Parallel Projection . . . . Step Proportion . . . . . . Examples . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
1 1 5 7 10 11 13
2 Rotations. Turns. Arc Steps 15 Definitions and Theorems of Rotation . . . . . . . . . . . . . . . . . . 15 Definitions of Turn and Arc Steps . . . . . . . . . . . . . . . . . . . . 17 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3 Quaternions Definitions and Theorem . Examples . . . . . . Multiplication . . . . . . . The Rotator q()q −1 . . . . Powers and Roots . . . . Representation of Vectors Examples . . . . . . Addition . . . . . . . . . . Formulas . . . . . . . . . Geometric Theorems . . . Examples . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
4 Equations of First Degree Scalar Equations, Plane and Straight Examples . . . . . . . . . . . . Nonions . . . . . . . . . . . . . . . Vector Equations, the Operator φ . . Linear Homogeneous Strain . . . . . Finite and Null Strains . . . . . . . . Solution of φρ = δ . . . . . . . . . . iv
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
23 23 24 25 26 27 28 30 31 36 38 41
Line . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
44 44 46 47 48 48 50 52
. . . . . . . . . . .
. . . . . . . . . . .
CONTENTS Derived Moduli. Latent Roots Latent Lines and Planes . . . . The Characteristic Equation . . Conjugate Nonions . . . . . . . Self-conjugate Nonions . . . . . Examples . . . . . . . . .
v . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
5 PROJECT GUTENBERG ”SMALL PRINT”
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
52 53 54 54 55 57
Chapter 1
Steps 1. Definition. A step is a given length measured in a given direction. E.g., 3 feet east, 3 feet north, 3 feet up, 3 feet north-east, 3 feet northeast-up, are steps. 2. Definition. Two steps are equal when, and only when, they have the same lengths and the same directions. E.g., 3 feet east, and 3 feet north, are not equal steps, because they differ in direction, although their lengths are the same; and 3 feet east, 5 feet east, are not equal steps, because their lengths differ, although their directions are the same; but all steps of 3 feet east are equal steps, whatever the points of departure. 3. We shall use bold-faced AB to denote the step whose length is AB, and whose direction is from A towards B.
Two steps AB, CD, are obviously equal when, and only when, ABDC is a parallelogram. 4. Definition. If several steps be taken in succession, so that each step begins where the preceding step ends, the step from the beginning of the first to the end of the last step is the sum of those steps.
1
CHAPTER 1. STEPS
2
√ E.g., 3 feet east + 3 feet north = 3 2 feet north-east = 3 feet north + 3 feet east. Also AB + BC = AC, whatever points A, B, C, may be. Observe that this equality between steps is not a length equality, and therefore does not contradict the inequality AB + BC > AC, just as 5 dollars credit + 2 dollars debit = 3 dollars credit does not contradict the inequality 5 dollars + 2 dollars > 3 dollars.
5. If equal steps be added to equal steps, the sums are equal steps. Thus if AB = A0 B0 , and BC = B0 C0 , then AC = A0 C0 , since the triangles ABC, A0 B 0 C 0 must be equal triangles with the corresponding sides in the same direction. 6. A sum of steps is commutative (i.e., the components of the sum may be added in any order without changing the value of the sum).
CHAPTER 1. STEPS
3
For, in the sum AB + BC + CD + DE + · · ·, let BC0 = CD; then since BCDC 0 is a parallelogram, therefore C0 D = BC, and the sum with BC, CD, interchanged is AB + BC0 + C0 D + DE + · · ·, which has the same value as before. By such interchanges, the sum can be brought to any order of adding. 7. A sum of steps is associative (i.e., any number of consecutive terms of the sum may be replaced by their sum without changing the value of the whole sum). For, in the sum AB + BC + CD + DE + · · ·, let BC, CD, be replaced by their sum BD; then the new sum is AB + BD + DE + · · ·, whose value is the same as before; and similarly for other consecutive terms. 8. The product of a step by a positive number is that step lengthened by the multiplier without change of direction. E.g., 2AB = AB + AB, which is AB doubled in length without change of direction; similarly 21 AB =(step that doubled gives AB) = (AB halved in length without change of direction). In general, mAB = m lengths AB measured in the direction AB; n1 AB = n1 th of length AB measured in the direction AB; etc. 9. The negative of a step is that step reversed in direction without change of length. For the negative of a quantity is that quantity which added to it gives zero; and since AB + BA = AA = 0, therefore BA is the negative of AB, or BA = −AB. • Cor. 1. The product of a step by a negative number is that step lengthened by the number and reversed in direction. For −nAB is the negative of nAB.
CHAPTER 1. STEPS
4
• Cor. 2. A step is subtracted by reversing its direction and adding it. For the result of subtracting is the result of adding the negative quantity. E.g., AB − CB = AB + BC = AC. 10. A sum of steps is multiplied by a given number by multiplying the components of the sum by the number and adding the products.
Let n·AB = A0 B0 , n·BC = BC0 ; then ABC, A0 B 0 C 0 are similar triangles, since the sides about B, B 0 are proportional, and in the same or opposite directions, according as n is positive or negative; therefore AC, A0 C 0 are in the same or opposite directions and in the same ratio; i.e., nAC = A0 C0 , which is the same as n(AB + BC) = nAB + nBC. This result may also be stated in the form: a multiplier is distributive over a sum. 11. Any step may be resolved into a multiple of a given step parallel to it; and into a sum of multiples of two given steps in the same plane with it that are not parallel; and into a sum of multiples of three given steps that are not parallel to one plane.
12. It is obvious that if the sum of two finite steps is zero, then the two steps must be parallel; in fact, if one step is AB, then the other must be equal to BA. Also, if the sum of three finite steps is zero, then the three steps must be parallel to one plane; in fact, if the first is AB, and the second is BC, then the third must be equal to CA. Hence, if a sum of steps on two lines that are not parallel (or on three lines that are not parallel to one
CHAPTER 1. STEPS
5
plane) is zero, then the sum of the steps on each line is zero, since, as just shown, the sum of the steps on each line cannot be finite and satisfy the condition that their sum is zero. We thus see that an equation between steps of one plane can be separated into two equations by resolving each step parallel to two intersecting lines of that plane, and that an equation between steps in space can be separated into three equations by resolving each step parallel to three lines of space that are not parallel to one plane. We proceed to give some applications of this and other principles of step analysis in locating a point or a locus of points with respect to given data (Arts. 13-20).
Centre of Gravity 13. The point P that satisfies the condition lAP + mBP = 0 lies upon the line AB and divides AB in the inverse ratio of l : m (i.e., P is the centre of gravity of a mass l at A and a mass m at B). The equation gives lAP = mPB; hence: AP, PB are parallel; P lies on the line AB; and AP : PB = m : l = inverse of l : m. If l : m is positive, then AP, PB are in the same direction, so that P must lie between A and B; and if l : m is negative, then P must lie on the line AB produced. If l = m, then P is the middle point of AB; if l = −m, then there is no finite point P that satisfies the condition, but P satisfies it more nearly, the farther away it lies upon AB produced, and this fact is expressed by saying that “P is the point at infinity on the line AB.” 14. By substituting AO + OP for AP and BO + OP for BP in lAP + mBP = 0, and transposing known steps to the second member, we find the point P with respect to any given origin O, viz., (a) (l + m)OP = lOA + mOB, where P divides AB inversely as l : m.
CHAPTER 1. STEPS
6
• Cor. If OC = lOA + mOB, then OC, produced if necessary, cuts AB in the inverse ratio of l : m, and OC is (l + m) times the step from O to the point of division. For, if P divide AB inversely as l : m, then by (a) and the given equation, we have OC = (l + m)OP. 15. The point P that satisfies the condition lAP + mBP + nCP = 0 lies in the plane of the triangle ABC; AP (produced) cuts BC at a point D that divides BC inversely as m : n, and P divides AD inversely as l : m + n (i.e., P is the center of gravity of a mass l at A, a mass m at B, and a mass n at C). Also the triangles P BC, P CA, P AB, ABC, are proportional to l, m, n, l + m + n.
The three steps lAP, mBP, nCP must be parallel to one plane, since their sum is zero, and hence P must lie in the plane of ABC. Since
CHAPTER 1. STEPS
7
BP = BD + DP, CP = CD + DP, the equation becomes, by making these substitutions, lAP + (m + n)DP + mBD + nCD = 0. This is an equation between steps on the two intersecting lines, AD, BC, and hence the resultant step along each line is zero; i.e., mBD + nCD = 0 (or D divides BC inversely as m : n), and
(a)
lAP + (m + n)DP = 0
(or P divides AD inversely as l : m + n). Also, we have, by adding lPD + lDP = 0 to (a), lAD + (l + m + n)DP = 0. Hence l : l + m + n = PD : AD = P BC : ABC, since the triangles P BC, ABC have a common base BC, (We must take the ratio of these triangles as positive or negative according as the vertices P , A lie on the same or opposite sides of the base BC, since the ratio PD : AD is positive or negative under those circumstances.) Similarly, P CA : ABC = m : l + m + n, and P AB : ABC = n : l + m + n. Hence, we have, P BC : P CA : P AB : ABC = l : m : n : l + m + n. 16. By introducing in lAP + mBP + nCP = 0 an origin O, as in Art. 14, we find (a) (l + m + n)OP = lOA + mOB + nOC, where P divides ABC in the ratio l : m : n. Note. As an exercise, extend this formula for the center of gravity P , of masses l, m, n, at A, B, C, to four or more masses.
Curve Tracing. Tangents. 17. To draw the locus of a point P that varies according to the law OP = tOA+ 21 t2 OB, where t is a variable number. (E.g., t = number of seconds from a given epoch.)
CHAPTER 1. STEPS
8
Take t = −2, and P is at D0 , where OD0 = −2OA + 2OB. Take t = −1, and P is at C 0 , where 1 OC0 = −OA + OB 2 Take t = 0, and P is at O. Take t = 1, and P is at C, where OC = OA + 12 OB. Take t = 2, and P is at D, where OD = 2OA + 2OB. It is thus seen that when t varies from -2 to 2, then P traces a curve D0 C 0 OCD. To draw the curve as accurately as possible, we find the tangents at the points already found. The method that we employ is perfectly general and applicable to any locus.
(a) To find the direction of the tangent to the locus at the point P corresponding to any value of t. Let P , Q be two points of the locus that correspond to the values t, t + h of the variable number. We have 1 OP = tOA + t2 OB, 2 1 OQ = (t + h)OA + (t + h)2 OB, 2
CHAPTER 1. STEPS
9
and therefore 1 PQ = OQ − OP = h OA + (t + h)OB . 2 Hence (dropping the factor h) we see that OA + (t + 12 h)OB is always parallel to the chord P Q. Make h approach 0, and then Q approaches P , and the (indefinitely extended) chord P Q approaches coincidence with the tangent at P . Hence making h = 0, in the step that is parallel to the chord, we find that OA + tOB is parallel to the tangent at P . Apply this result to the special positions of P already found, and we have: D0 A0 = OA − 2OB = tangent at D0 ; C0 S = OA − OB = tangent at C 0 ; OA = OA + 0 · OB = tangent at O; SO = OA + OB = tangent at C; AD = OA + 2OB = tangent at D. This is the curve described by a heavy particle thrown from O with velocity represented by OA on the same scale in which OB represents an acceleration of 32 feet per second per second downwards. For, after t seconds the particle will be displaced a step t · OA due to its initial velocity, and a step 21 t2 · OB due to the acceleration downwards, so that P is actually the step OP = tOA + 12 t2 · OB from O at time t. Similarly, since the velocity of P is increased by a velocity represented by OB in every second of time, therefore P is moving at time t with velocity represented by OA + tOB, so that this step must be parallel to the tangent at P . 18. To draw the locus of a point P that varies according to the law OP = cos(nt + e) · OA + sin(nt + e) · OB, where OA, OB are steps of equal length and perpendicular to each other, and t is any variable number. With centre O and radius OA draw the circle ABA0 B 0 . Take arc AE = e radians in the direction of the quadrant AB (i.e. an arc of e radii of the circle in length in the direction of AB or AB 0 according as e is positive or negative). Corresponding to any value of t, lay off arc EP = nt radians in the direction of the quadrant AB. Then arc AP = nt + e radians. Draw LP perpendicular to OA at L. Then according to the definitions of the trigonometric functions of an angle we have, cos(nt + e) = OL/OP,
sin(nt + e) = LP /OP.1
Hence we have for all values of t, OL = cos(nt + e)OA,
LP = sin(nt + e)OB,
1 Observe the distinctions: OL, a step; OL, a positive or negative length of a directed axis; OL, a length.
CHAPTER 1. STEPS
10
and adding these equations, we find that OP = cos(nt + e)OA + sin(nt + e)OB. Hence, the locus of the required point P is the circle on OA, OB as radii.
Let t be the number of seconds that have elapsed since epoch. Then, at epoch, t = 0, and P is at E; and since in t seconds P has moved through an arc EP of nt radians, therefore P moves uniformly round the circle at the rate of n radians per second. Its velocity at time t is therefore represented by n times that radius of the circle which is perpendicular to OP in the direction of its motion, or by OP0 = nOQ, where arc π P Q = π2 radians. Hence, since there arc AQ = (nt + e + 2 ) radians, fore OP0 = n cos nt + e + π2 · OA + sin nt + e + π2 · OB . The point P 0 also moves uniformly in a circle, and this circle is the hodograph of the motion. The velocity in the hodograph (or the acceleration of P ) is similarly OP00 = n2 PO.
Parallel Projection 19. If OP = xOA + yOB, OP0 = xOA + yOB0 , where x, y vary with the arbitrary number t according to any given law so that P , P 0 describe
CHAPTER 1. STEPS
11
definite loci (and have definite motions when t denotes time), then the two loci (and motions) are parallel projections of each other by rays that are parallel to BB 0 , For, by subtracting the two equations we find PP0 = yBB0 , so that P P 0 is always parallel to BB 0 ; and as P moves in the plane AOB and P 0 moves in the plane AOB 0 , therefore their loci (and motions) are parallel projections of each other by rays parallel to BB 0 . The parallel projection is definite when the two planes coincide, and may be regarded as a projection between two planes AOB, AOB 0 , that make an indefinitely small angle with each other.
20. The motion of P that is determined by OP = cos(nt + e)OA + sin(nt + e)OB is the parallel projection of uniform circular motion. For, draw a step OB0 perpendicular to OA and equal to it in length. Then, by Art. 18, the motion of P 0 determined by OP0 = cos(nt + e)OA + sin(nt + e)OB0 is a uniform motion in a circle on OA, OB0 as radii; and by Art. 19 this is in parallel perspective with the motion of P .
Step Proportion 21. Definition. Four steps AC, AB, A0 C0 , A0 B0 are in proportion when the first is to the second in respect to both relative length and relative direction as the third is to the fourth in the same respects.
CHAPTER 1. STEPS
12
This requires, first, that the lengths of the steps are in proportion or AC : AB = A0 C 0 : A0 B 0 ; and secondly, that AC deviates from AB by the same plane angle in direction and magnitude that A0 C0 deviates from A0 B0 . Hence, first, the triangles ABC, A0 B 0 C 0 are similar, since the angles A, A0 are equal and the sides about those angles are proportional; and secondly, one triangle may be turned in its plane into a position in which its sides lie in the same directions as the corresponding sides of the other triangle. Two such triangles will be called similar and congruent triangles, and corresponding angles will be called congruent angles. 22. We give the final propositions of Euclid, Book V., as exercises in step proportion. • (xi.) If four steps are proportionals, they are also proportionals when taken alternately. • (xii.) If any number of steps are proportionals, then as one of the antecedents is to its consequent, so is the sum of the antecedents to the sum of the consequents. • (xiii.) If four steps are proportionals, the sum (or difference) of the first and second is to the second as the sum (or difference) of the third and fourth is to the fourth. • (xiv.) If OA : OB = OP : OQ and OB : OC = OQ : OR, then OA : OC = OP : OR.
CHAPTER 1. STEPS
13
• (xv.) If OA : OB = OC : OD and OE : OB = OF : OD, then OA + OE : OB = OC + OF : OD. • (xvi.) If OA : OB = OB : OX = OC : OD = OD : OY, then OA : OX = OC : IO.
Examples We shall use i, j, k, as symbols for unit length east, unit length north, and unit length up, respectively. 1. Mark the points whose steps from a given point are i + 2j, −3i − j. Show that the step from the first point to the second is −4i − 3j, and that the length is 5. 2. Show that the four points whose steps from a given point are 2i + j, 5i + 4j, 4i + 7j, i + 4j are the angular points of a parallelogram. Also determine their centre of gravity, with weights 1, 1, 1, 1; also with weights 1, 2, 3, 4; also with weights 1, −2, 3, −4. 3. If OA = i + 2j, OB = 4i + 3j, OC = 2i + 3j, OD = 4i + j, find CD as sums of multiples of CA, CB, and show that CD bisects AB. 4. If OP = xi + yj, OP0 = x0 i + y 0 j, then PP0 = (x0 − x)i + (y 0 − y)j and 2 P P 0 = (x0 − x)2 + (y 0 − y)2 . 5. Show that AB is bisected by OC = OA + OB, and trisected by OD = 2OA + OB, OE = OA + 2OB, and divided inversely as 2 : 3 by OF = 2OA + 3OB. 6. Show that AA0 + BB0 = 2MM0 , where M M 0 are the middle points of AB, A0 B 0 , respectively. 7. Show that 2AA0 + 3BB0 = (2 + 3)CC0 , where C, C 0 are the points that divide AB, A0 B 0 , inversely as 2 : 3. Similarly, when 2, 3 are replaced by l, m. 8. Show that the point that divides a triangle into three equal triangles is the intersection of the medial lines of the triangle. 9. Show that the points which divide a triangle into triangles of equal magnitude, one of which is negative (the given triangle being positive), are the vertices of the circumscribing triangle with sides parallel to the given triangle. 10. If a, b, c are the lengths of the sides BC, CA, AB of a triangle, show that 1 AC ± 1c AB (drawn from A) are interior and exterior bisectors of the angle b A; and that when produced they cut the opposite side BC in the ratio of the adjacent sides. lines vertices point P 11. The that join the of a triangle ABC to any in points sides line p +1 its plane divide the sides BC, CA, AB in ratios whose product is ; and −1 lines from vertices meet in a point. conversely the that so divide the sides points on sides lie in a line. 12. Prove by Exs. 10, 11, that the three interior bisectors of the angles of a triangle (also an interior and two exterior bisectors) meet in a point; and that the three exterior bisectors (also an exterior and two interior bisectors) meet the sides in colinear points.
CHAPTER 1. STEPS
14
13. Determine the locus (and motion) of P , given by OP = OA + tOB; also of OP = (1 + 2t)i + (3t − 2)j. 14. Compare the loci of P determined by the following pairs of step and length equations: AP = 2 east,
AP = 2;
AP + BP = CD,
AP = 2BP,
AP = 2BP ;
AP + BP = CD
15. Draw, by points and tangents, the locus of P determined by each of the following values of OP, in which x is any number: 1 2 1 1 xi + x2 j; xi + j; xi + x3 j; xi + ( x3 − x2 + 2)j; 2 x 3 3 p p 8 1 xi + 2 j; xi + 4 − x2 j; xi + 4 − x2 j. x +4 2 16. Take three equal lengths making angles 120◦ with each other as projections of i, j, k, and construct by points the projection of the locus of P , where OP = 2(cos x·i + sin x·j) + x·k, x varying from 0 to 2π. Show that this curve is one turn of a helix round a vertical cylinder of altitude 2π, the base being a horizontal circle of radius 2 round O as centre. 17. A circle rolls inside a fixed circle of twice its diameter; show that any point of the plane of the rolling circle traces a parallel projection of a circle. 18. A plane carries two pins that slide in two fixed rectangular grooves; show that any point of the sliding plane traces a parallel projection of a circle. 19. OACB is a parallelogram whose sides are rigid and jointed so as to turn round the vertices of the parallelogram; AP C, BCQ are rigid similar and congruent triangles. Show that AC : AP = BQ : BC = OQ : OP, and that therefore P , Q trace similar congruent figures when O remains stationary (21, 22, xii.). [See cover of book.] 20. If the plane pencil OA, OB, OC, OD is cut by any straight line in the points P , Q, B, S, show that the cross-ratio (P R : RQ) : (P S : SQ) is constant for all positions of the line. [OC = lOA + mOB = lxOP + myOQ gives P R : RQ = my : lx]. 21. Two roads run north, and east, intersecting at O. A is 60 feet south of O, walking 3 feet per second north, B is 60 feet west of O, walking 4 feet per second east. When are A, B nearest together, and what is B’s apparent motion as seen by A? 22. What is B’s motion relative to A in Ex. 21 if B is accelerating his walk at the rate of 3 inches per second per second ? 23. In Ex. 21, let the east road be 20 feet above the level of the north road; and similarly in Ex. 22. 24. A massless ring P is attached to several elastic strings that pass respectively through smooth rings at A, B, C, · · · and are attached to fixed points A0 , B 0 , C 0 , · · · such that A0 A, B 0 B, C 0 C, · · · are the natural lengths of the strings. The first string has a tension l per unit of length that it is stretched (Hooke’s law), the second a tension m, the third a tension n, etc. Find the resultant force on P and its position of equilibrium. 25. The same as Ex. 24, except that the ring has a mass w.
Chapter 2
Rotations. Turns. Arc Steps 23. Definitions of Rotation A step is rotated when it is revolved about an axis through its initial point as a rigid length rigidly attached to the axis. The step describes a conical angle about the axis except when it is perpendicular to the axis.
If a rotation through a diedral angle of given magnitude and direction in space be applied to the radii of a sphere of unit radius and centre O, the sphere is rotated as a rigid body about a certain diameter P P 0 as axis, and a plane through O perpendicular to the axis intersects the sphere in the equator of the rotation. Either of the two directed arcs of the equator from the initial position A to the final position A0 of a point of the rotated sphere that lies on the equator is the arc of the rotation. If these two arcs he bisected at 15
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS _
16 _
L, M respectively, then the two arcs are 2AL, 2AM respectively, and _
_
AL, AM are supplementary arcs in opposite directions, each less than a semicircle. When these half-arcs are 0◦ and 180◦ respectively, they represent a rotation of the sphere into its original position, whose axis and equator are indeterminate, so that such arcs may be measured on any great circle of the sphere without altering the corresponding rotation. 24. A rotation is determined by the position into which it rotates two given non-parallel steps. For let the radii OB, OC rotate into the radii OB0 , OC0 . Any axis round which OB rotates into OB0 must be equally inclined to these radii; i.e., _
it is a diameter of the great circle P KL that bisects the great arc BB 0 at right angles.
E.g., OK, OL, OP , · · · are such axes. Similarly, the axis that rotates OC into OC0 must be a diameter of the great circle P N that bisects the great _
arc CC 0 at right angles. Hence there is but one axis round which OB, OC rotate into OB0 , OC0 ; viz., the intersection OP of the planes of these two bisecting great circles: the equator is the great circle whose plane is perpendicular to this axis, and the arcs of the rotation are the intercepts on the equator by the planes through the axis and either B, B 0 or C, C 0 . [When the two bisecting great circles coincide (as when C, C 0 lie on BP , B 0 P ), then their plane bisects the diedral angle BC − O − B 0 C 0 , whose edge OP is the only axis of rotation.] _
_
Note. Since BC, B 0 C 0 may be any two positions of a marked arc on the surface of the sphere, we see that any two positions of the sphere with centre fixed determine a definite rotation of the sphere from one position to the other.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
17
25. A marked arc of a great circle of a rotating sphere makes a constant angle with the equator of the rotation. For the plane of the great arc makes a constant angle both with the axis and with the equator of the rotation. _
_
26. If the sphere O be given a rotation 2A0 C followed by a rotation 2CB0 , the _
resultant rotation of the sphere is 2A0 B0 .
_
_
_
For produce the arcs A0 C, B0 C to A1 , B 0 respectively, making CA1 = _
_
_
A0 C, B 0 C = CB0 . Then the spherical triangles A0 B0 C, A1 B 0 C are equal, since the corresponding sides about the equal vertical angles at C are by _
_
construction equal. Therefore the sides A0 B0 , B 0 A1 are equal in length, and the corresponding angles A0 , A1 and B0 , B 0 are equal. Therefore, by _
_
Art. 25, if a marked arc AB of the sphere coincide initially with A0 B0 , the first rotation _ 0
_ 2A0 C
=
_ A0 A1
_
will bring AB into the position _ 0
_
_ A1 B1 _
on
B A1 produced, and the second rotation 2CB0 = B B1 will bring AB into _
_
_
_
the position A2 B2 on A0 B0 produced, where B0 A2 = A0 B0 . Hence the _
_
resultant rotation of the sphere is 2A0 B0 = A0 A2 . Note. This theorem enables one to find the resultant of any number of successive rotations, by replacing any two successive rotations by their resultant, and so on until a single resultant is found.
27. Definitions of Turn A step is turned when it is made to describe a plane angle round its initial point as centre.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
18
If a turn through a plane angle of given magnitude and direction in space be applied to the radii of the sphere O, it turns the great circle that is parallel to the given plane angle as a rigid circle, and does not affect the other radii of the sphere. E.g., only horizontal radii can be turned through a horizontal plane angle. The circle that is so turned is the great circle of the turn. A directed arc of the great circle of a turn from the initial position A to the final position B of a point on the great circle, and less than a semi-circumference, is the arc of the turn. When this arc is 0◦ or 180◦ , it represents a turn that brings a step back to its original position or that reverses it; and since such turns may take place in any plane with the same results, therefore such arcs may be measured on any great circle of the sphere without altering their corresponding turns. The axis of a turn is that radius of the sphere O which is perpendicular to its great circle and lies on that side of the great circle from which the arc of the turn appears counter-clockwise. 28. A turn is determined by the position into which it displaces any given step. For, let the radius OA turn into the radius OB. Then, the great circle _
O − AB must be the great circle of the turn, and AB, the arc of the turn. _
_
29. Definitions. The resultant of two successive turns AB, BC is the turn _
AC.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
19
When the arc of the turns are not given with the first ending where the second begins, each arc may be moved as a rigid arc round its great circle until they do so end and begin, without altering their turning value. When the two great circles are not the same, then the common point of the two arcs must be one or the other point of intersection (B, B 0 ) of the two great circles. The figure shows that the same resultant is found from either of these points. ARC STEPS _
We may call the great arc AB the arc step from A to B on the surface of the sphere; and call two arc steps equal when they are arcs of the same _
great circle of the same length and direction; and call AC the sum of _
_
AB, BC or the sum of any arc steps equal to these. The half-arc of a resultant rotation is thus the sum of the half-arcs of its components, and the arc of a resultant turn is the sum of the arcs of the components. The sum of several arcs is found by replacing any two successive arcs of the sum by their sum, and so on, until a single sum is found. An arc of 0◦ or 180◦ may be measured on any great circle without altering its value as the representative of a half-rotation, a turn, or an arc step. 30. The resultant of two successive rotations or turns (i.e., the sum of two arc steps) is commutative only when the arcs are cocircular. _
_
For let the half-arcs of the rotations, or the arcs of the turns, be AB = BA0 , _
_
_
_
_
_
and C 0 B = BC; then the sums AB + BC, C 0 B + BA0 in opposite orders _
_
are respectively AC, C 0 A0 ; and from the figure those arcs are equal when, and only when, the given arcs are cocircular.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
20
• Cor. 1. An arc of 0◦ or 180◦ is commutative with any other arc. For it may be taken cocircular with the other arc. • Cor. 2. The magnitudes of the sums of two arcs in opposite orders are equal. For ABC, A0 BC 0 are equal spherical triangles by construction, and _
_
therefore AC, C 0 A0 are equal in length. 31. A sum of successive arc steps is associative.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
21
For, consider first three arcs upon the great circles LQ0 , Q0 R, RL. If the arcs are such as to begin and end successively, the proof is the same as for _
_
_
_
step addition, e.g., in the sum AQ0 + Q0 R + RB = AB, the first two may _
_
be replaced by their sum AR, or the second and third by their sum Q0 B without altering the whole sum. In the more general case when the three arcs are _ _ _ _ _ _ AQ0 = S 0 P 0 , Q0 Q = R0 R, RB = P T , _
_
_
the sum of the first two is AQ = SP , whose sum with the third is ST ; _ 0
_ 0 0
and the sum of the second and third is R B = P T , whose sum with the _
_
_
first is S 0 T 0 ; and we must prove that ST , S 0 T 0 are equal arcs of the same great circle in the same direction. [Observe that in the construction P is determined as the intersection of QA and RB, and P 0 as the intersection of Q0 A and R0 B.] Let the three given arcs be the half-arcs of successive rotations of the _
_
sphere O. Then by Art. 26, the rotation 2AQ = 2SP gives the sphere the _
same displacement as the first and second rotations, so that 2ST gives the sphere the same displacement as the three rotations. Similarly, the rota_
_
tion 2R0 B = 2P 0 T gives the sphere the same displacement as the second _
and third rotations, so that 2S 0 T 0 gives the sphere the same displacement _
_
as the three rotations. Hence ST , S 0 T 0 are arcs of the same great circle, and either equal (and in the same direction) or supplementary (and in opposite directions), since they are half-arcs of the same rotation. This is true wherever Q may be. Suppose that Q is slightly displaced towards _
_
R; then ST , S 0 T 0 are slightly displaced, and if equal at first, they must remain equal, since a slight change in each of two equal arcs could not change them to supplementary arcs in opposite directions.1 Hence by _
_
moving Q continuously towards R and finding how the arcs ST , S 0 T 0 are related when Q reaches R, we find how they are related for any position of Q, since there is no change in the relation when Q is moved continuously. But when Q is at R, it was shown above that both arcs were equal; _
_
therefore ST , S 0 T 0 are always equal. So, in general, for a sum of any number of successive arcs, any way of forming the sum by replacing any two successive terms by their sum and so on, must give a half-arc of the resultant of the rotations through double each of the given arcs. Hence any two such sums are either equal or opposite supplementary arcs of the same great circle; and since by continuous changes of the component arcs, they may be brought so that each begins 1 When both arcs are nearly 90◦ , a slight change in each could change them from equals to supplements in the same direction.
CHAPTER 2. ROTATIONS. TURNS. ARC STEPS
22
where the preceding arc ends, in which position the two sums are equal, therefore they are always equal. • Cor. 1. An arc of 0◦ or 180◦ may have any position in a sum. [Art. 30, Cor. 1.] • Cor. 2. The magnitude of a sum of arcs is not changed by a cyclic change in the order of its terms. _
_
_
_
_
_
For (AB + CD + · · · ) + HK and HK + (AB + CD + · · · ) have equal magnitudes. [Art. 30, Cor. 2.]
EXAMPLES _
_
_
_
1. Show that 2(AB + BC) and 2AB + 2BC are in general unequal. 2. If (2, 30◦ ) denote a turn of 30◦ counter-clockwise in the plane of the paper and a doubling, and (3, −60◦ ) denote a turn of 60◦ clockwise in the plane of the paper and a trebling, express the resultant of these two compound operations (versi-tensors) in the same notation. 3. Find the resultant of (2, 30◦ ), (3, 60◦ ), (4, −120◦ ), (1, 180◦ ). 4. Show that either (2, −60◦ ) or (2, 120◦ ) taken twice have the resultant (4, −120◦ ). 5. Would you consider the resultants of versi-tensors as their sums or their products, and why? 6. Let the base QR of a spherical triangle P QR slide as a rigid arc round its fixed great circle, and let the great circles QP , RP , always pass through fixed points A, B respectively. Show that if points S, T lie on the great circles QP , RP so _
_
_
_
_
as always to keep P S = QA and P T = RB, then the arc ST is an arc of fixed _
length and direction that slides around a fixed great circle as QR slides round its fixed great circle. [Let P 0 , Q0 , R0 , S 0 , T 0 , be given positions of P , Q, R, S, T , and use Art. 31 and figure.] 7. Show that the locus of the radius OP in Ex. 6 is an oblique circular cone of which OA, OB are two elements, and that the fixed great circles QR, ST are parallel to its circular sections. [Draw a fixed plane parallel to OQR and cutting the radii OA, OB, in the fixed points A0 , B 0 , and cutting OP in the variable point P 0 , and show that P 0 describes a circle in this plane through the fixed points A0 , B 0 ; similarly, for a fixed plane parallel to OST .] Note.—The locus of P on the surface of the sphere is called a spherical conic (the intersection of a sphere about the vertex of a circular cone as centre with the surface of the cone); and the great circles QR, ST (parallel to the circular sections of the cone) are the cyclic great circles of the spherical conic. The above properties of a spherical conic and its cyclic great circles become properties of a plane conic and its asymptotes when the centre O of the sphere is taken at an indefinitely great distance. 8. State and prove Ex. 6 for a plane, and construct the locus of P .
Chapter 3
Quaternions 32. Definitions. A quaternion is a number that alters a step in length and direction by a given ratio of extension and a given turn. E.g., in the notation of Ex. 2, II, (2, 30◦ ), (2, −60◦ ) are quaternions. Two quaternions are equal when, and only when, their ratios of extension are equal and their turns are equal. A tensor is a quaternion that extends only; i.e., a tensor is an ordinary positive number. Its turn is 0◦ in any plane. A versor or unit is a quaternion that turns only. E.g., 1, −1 = (1, 180◦ ), (1, 90◦ ), (1, 30◦ ), are versors. A scalar is a quaternion whose product lies on the same line or “scale” as the multiplicand; i.e., a scalar is an ordinary positive or negative number. Its turn is 0◦ or 180◦ in any plane. A vector is a quaternion that turns 90◦ . E.g., (2, 90◦ ), (1, −90◦ ), are vectors. 33. Functions of a Quaternion q. The tensor of q, or briefly T q, is its ratio of extension. E.g., T 2 = 2 = T (−2) = T (2, 30◦ ). The versor of q (U q) is the versor with the same arc of turn as q. E.g., U 2 = 1,
U (−2) = −1,
U (2, 30◦ ) = (1, 30◦ ).
The arc, angle, axis, great circle, and plane of q, are respectively the arc, angular magnitude, axis, great circle, and plane of its turn. E.g., arc(2, 30◦ ) is a counter-clockwise arc of 30◦ of unit radius in the plane of the paper, and arc(2, −30◦ ) is the same arc oppositely directed; ∠(2, 30◦ ) = ∠(2, −30◦ ) = 30◦ = π6 radians; axis(2, 30◦ ) is a unit length perpendicular to the plane of the paper directed towards the reader, and axis(2, −30◦ ) is the same length oppositely directed; etc.
23
CHAPTER 3. QUATERNIONS
24
If qOA = OB, and if L be the foot of the perpendicular from B upon the line OA, then OL, LB are called the components of q’s product respectively parallel and perpendicular to the multiplicand; also, the projections of OB parallel and perpendicular to OA. The scalar of q (Sq) is the scalar whose product equals the component of q’s product parallel to the multiplicand; viz., Sq · OA = OL. √ √ E.g., S(2, 30◦ ) = 3, S(2, 150◦ ) = − 3. The vector of q (V q) is the vector whose product equals the component of q’s product perpendicular to the multiplicand; viz., V q · OA = LB. √ E.g., V (2, 30◦ ) = (1, 90◦ ) = V (2, 150◦ ), V (2, −60◦ ) = ( 3, −90◦ ). The reciprocal of q (1/q or q −1 ) is the quaternion with reciprocal tensor and reversed turn. E.g., (2, 30◦ )−1 = ( 21 , −30◦ ). The conjugate of q(Kq) is the quaternion with the same tensor and reversed turn. E.g., K(2, 30◦ ) = (2, −30◦ ). 34. From the above diagram and the definitions of the cosine and sine of an angle, we have (a) (b)
OL OB OL OL = = · = T q · cos ∠q OA OA OA OB LB OB LB TV q = = · = T q · sin ∠q OA OA OB
Sq =
Note. Arc V q is a quadrant on the great circle of q in the direction of arc q.
Examples 1. If equal numbers multiply equal steps, the products are equal; and if they multiply unequal steps, the products are unequal. 2. If the products of two steps by equal numbers are equal, then the two steps are equal; and if the products of two equal steps by two numbers are equal, then the numbers are equal.
CHAPTER 3. QUATERNIONS
25
3. If several steps be multiplied by equal numbers, then any product is to its multiplicand as any other product is to its multiplicand. 4. If two steps be multiplied by reciprocal numbers, then corresponding products and multiplicands are reciprocally proportional. 5. Construct the following products, where OA is a unit step to the right in the plane of the paper, and determine the functions of each multiplier that are defined in Art. 33. (a) 2 ·√ OA = OL, (4, 60◦ ) · OA√= OB, (4, −60◦ ) · OA = OB0 , (2 3, 90◦ ) · OA = OM, (2 3, −90◦ ) · OA = OM0 , (1, 60◦ ) · OA = OB1 , (1, −60◦ ) · OA = OB01 , (1, 90◦ ) · OA = OM1 , (1, −90◦ ) · OA = OM01 . (b) The same as (a) with 120◦ in the place of 60◦ . 6. Show that SSq = Sq, SV q = 0, V Sq = 0, V V q = V q, SKq = KSq = Sq, V Kq = KV q, U Sq = ±1, U T q = 1 = T U q.
Multiplication 35. Definition. The product of two or more numbers is that number whose extension and turn are the resultants of the successive extensions and turns of the factors (beginning with the right-hand factor). E.g., if rOA = OB, qOB = OC, pOC = OD, then we have pqr · OA = pqOB = pOC = OD. 36. The product is, however, independent of whether a step OA can be found or not, such that each factor operates upon the product of the preceding factor; i.e., we have by definition, (a) T (· · · pqr) = · · · T p · T q · T r. (b) arc (· · · pqr) = arc r + arc q + arc p + · · · . 37. The product of a tensor and a versor is a number with that tensor and versor; and conversely, a number is the product of its tensor and its versor. For if n be a tensor, and q 0 a versor, then nq 0 turns by the factor q 0 and extends by the factor n, and vice versa for q 0 n; hence either of the products, nq 0 , q 0 n, is a quaternion with tensor n and versor q0. Similarly, q = T q · U q = U q · T q. 38. Any successive factors of a product may be replaced by their product without altering the value of the whole product; but in general such factors can be changed in order without altering the value of the product only when those factors are cocircular. For replacing successive factors by their product does not alter the tensor of the whole product by Art. 36(a), nor the arc of the product by Art. 31, 36(b); but by Art. 30 the arc of the product is altered if two factors be interchanged except when those factors are cocircular.
CHAPTER 3. QUATERNIONS
26
• Cor. 1. A scalar factor may have any position in the product without altering the value of the product. [Art. 31, Cor. 1.] • Cor. 2. The angle of a product is not altered by a cyclic change in the order of the factors. [Art. 31, Cor. 2.] • Cor. 3. The scalar, and the tensor of the vector, of a product are not altered by a cyclic change in the order of the factors. [Art. 34, a, b.] 39. The product of two numbers with opposite turns equals the product of the tensors of the numbers; and conversely if the product of two numbers is a tensor, then the turns of the factors are opposites. [36 a, b.] • Cor. 1. The product of two conjugate numbers equals the square of their tensor; and if the product of two numbers with equal tensors is a tensor, then the two numbers are conjugates. • Cor. 2. The conjugate of a product equals the product of the conjugates of the factors in reverse order. For (pqr)(Kr · Kq · Kp) = (T p)2 · (T q)2 · (T r)2 since rKr = (T r)2 , may have any place in the product, and may be put first; and then (qKq) = (T q)2 , may be put second, and then (pKp) = (T p)2 . [Cor. 1, 38 Cor. 1.] Hence, K(pqr) = Kr · Kq · Kp. [Cor. 1.] • Cor. 3. The product of two reciprocal numbers is unity; and conversely, if the product of two numbers with reciprocal tensors is unity, then the numbers are reciprocals. • Cor. 4. The reciprocal of a product equals the product of the reciprocals of the factors in reverse order. For (pqr)(r−1 q −1 p−1 ) = 1. 40. The square of a vector is −1 times the square of its tensor; and conversely, if the square of a number is a negative scalar, then the number is a vector. [36, a, b.] • Cor. 1. The conjugate of a vector is the negative vector. [39 Cor. 1.] • Cor. 2. The conjugate of a product of two vectors is the product of the same vectors in reverse order. [Art. 39, Cor. 2.] • Cor. 3. The conjugate of a product of three vectors is the negative of the product of the same vectors in reverse order. [Art. 39, Cor. 2.]
CHAPTER 3. QUATERNIONS
27
The Rotator q()q −1 41. We may consider the ratio of two steps as determining a number, the antecedent being the product and the consequent the multiplicand of the number; viz., OB/OA determines the number r such that rOA = OB. By Art. 21, equal step ratios determine equal numbers. If the several pairs of steps that are in a given ratio r be given a rotation whose equatorial arc is 2 arc q, they are still equal ratios in their new positions and determine a new number r0 that is called the number r rotated through 2 arc q. In other words, the rotation of r produces a number with the same tensor as r, and whose great circle and arc are the rotated great circle and arc of r. 42. The number r rotated through 2 arc q is the number qrq −1 . For, 1st, T qrq −1 = T q · T r(T q)−1 = T r.
2d, let A be an intersection of the great circle of r with the great circle of q and construct _
_
AB = BA0 = arc q,
_
AC = arc r,
and _
_
C 0 B = BC = arc rq −1 ; then _
_
C 0 A0 = A0 C 00 = arc qrq −1 . But by construction, the spherical triangles ABC, A0 BC 0 are equal, and _
_
_
therefore AC and C 0 A0 (= A0 C 00 ) are arcs of equal length, and the corre_
sponding angles at A, A0 are equal. Hence, when arc r(= AC) is rotated _
_
through 2 arc q(= AA0 ), it becomes arc qrq −1 (= A0 C 00 ).
CHAPTER 3. QUATERNIONS
28
Powers and Roots 43. An integral power, q n = q · q · q · · · to n factors, is determined by the equations, (a) T · q n = T q · T q · T q · · · = (T q)n . (b) arc q n = arc q + arc q + arc q · · · = n arc q ± (whole circumferences). 1 To find q n , the number whose nth power is q, we have, by replacing 1 q by q n in (a), (b), 1
1
1
(c) T q = (T · q n )n or T · q n = (T q) n 1
1
(d) Arc q = n arc q n ± whole circumferences, or, arc q n = n1 (arc q± whole circumferences) = n1 arc q + m n circumferences ± whole circumferences), where m = 0, 1, 2, 3, · · · n − 1, successively. There are therefore n nth roots of q whose tensors are all equal and whose arcs lie on the great circle of q. When the base is a scalar, its great circle may be any great circle, so that there are an infinite number of quaternion nth roots of a scalar. On this account, the roots as well as the powers of a scalar are limited to scalars. By ordinary algebra, there are n such nth roots, real and imaginary. There are also imaginary nth √ roots of q besides the n real roots found above; i.e., roots of the form a + b −1, where a, b are real quaternions.
Representation of Vectors 44. Bold-face letters will be used as symbols of vectors only. In particular, i, j, k will denote unit vectors whose axes are respectively a unit length east, a unit length north, and a unit length up. More generally we shall use the step AB to denote the vector whose axis is a unit length in the direction of AB, and whose tensor is the numerical length of AB (= AB: unit length). This use of a step AB as the symbol of a vector is analogous to the use of AB to represent a tensor (AB: unit length), or of AB to represent a positive or negative scalar, according as it is measured in or against the direction of its axis of measurement. In none of these cases is the concrete quantity an absolute number; i.e., the value of the number that it represents varies with the assumed unit of length. When desirable, we distinguish between the vector OA and the step OA by enclosing the vector in a parenthesis. 45. If q(OA) = (OB), then q · OA = OB, and conversely. The tensor of q in either equation is OB : OA. It is therefore only necessary to show that the arc of q in one equation equals the arc of q in the
CHAPTER 3. QUATERNIONS
29
other equation in order to identify the two numbers that are determined by these two equations as one and the same number.
Draw the sphere of unit radius and centre O, cutting OA, OB in A0 , B 0 ; _
then A0 B 0 is the arc of q in the second equation. Draw the radius OL _
perpendicular to the plane OA0 B 0 on the counter-clockwise side of A0 B 0 , _
and draw counter-clockwise round OA0 , OB 0 as axes the quadrants LM , _
LN respectively; then these are the arcs of (OA), (OB) respectively, _
_
_
_
and since LM + M N = LN , therefore M N is the arc of q in the first _
_
equation. But since LM , LN are quadrants, therefore the plane OM N is perpendicular to OL, and must therefore coincide with the plane OA0 B 0 , _
which is by construction also perpendicular to OL. Hence M N lies on the _
great circle of A0 B 0 , and by the construction of the figure, it must, when _
advanced 90◦ on that great circle, coincide with A0 B 0 . Hence the theorem. Note. This theorem shows that a number extends and turns vectors into vectors in the same way that it extends and turns steps into steps. Moreover, when the vector is not perpendicular to the axis of the multiplier, there is no resulting vector, since in the case of the corresponding step there is no resulting step. In the case of a vector multiplicand, that is oblique to the axis of q, the product is an actual quaternion that is not a vector, while in the case of the corresponding step multiplicand the product belongs to that class of products in which the multiplicand does not admit of the operation of the multiplier, as √ in 2 universities, -2 countries, etc.
• Cor. 1. The product of two vectors is a vector when, and only when, the factors are perpendicular to each other; the product is perpendicular to both factors; and its length (its tensor) is equal to the area of the rectangle on the lengths of the factors.
CHAPTER 3. QUATERNIONS
30
Note. The direction of the product OA·OB = OC is obtained by turning OB about OA as axis through a counter-clockwise right angle; thus OC lies on that side of the plane OAB from which the right angle AOB appears counter-clockwise.
• Cor. 2. The product of two perpendicular vectors changes sign when the factors are interchanged. (OB · OA = OC0 = −OC.)
• Cor. 3. The condition that α is perpendicular to β is that αβ = vector, or Sαβ = 0. 46. If AB, CD are parallel, then AB · CD = CD · AB = −AB·CD, a scalar; and conversely, the product of two vectors is a scalar only when they are parallel. Since the axes of the vectors AB, CD are parallel, therefore their product is commutative. When the vectors are in the same direction, then each turns 90◦ in the same direction, the resultant turn is 180◦ , and the product is negative; and when the vectors are in opposite direction, their turns are in opposite directions, the resultant turn is 0◦ , and the product is positive. This is just the opposite of the product of the corresponding scalars AB, CD, which is positive when the scalars are in the same direction (or both of the same sign), and negative when the scalars are in opposite directions; i.e., AB·CD = −AB · CD. Conversely, the product AB, CD can be a scalar only when the resultant of their two turns of 90◦ each is a turn of 0◦ or 180◦ ; i.e., only when the turns are cocircular, and therefore their axes parallel. • Cor. The condition that α is parallel to β is αβ = scalar, or V αβ = 0.
Examples 1. Prove by diagram that (pq)2 and p2 q 2 are in general unequal. 2. Find the 2d, 3d, 4th, 5th, 6th powers of (2, 90◦ ), (2, −60◦ ).
CHAPTER 3. QUATERNIONS
31
3. Find the square roots and cube roots of (4, 30◦ ), (8, −120◦ ). 1
1
6
(a) Find the values of [(2, 50◦ )6 ] 3 , [(2, 50◦ ) 3 ]6 , and (2, 50◦ ) 3 . 4. What numbers are represented by 2 feet, 2 feet east, the unit of length being a foot, a yard, an inch? 5. Show that i2 = j2 = k2 = ijk = −1; jk = i = −kj; ki = j = −ik; ij = k = −ji. 6. Let e(AB) denote the versor that turns counter-clockwise round the axis AB through an arc that is formed by bending the length AB into an arc of unit radius. Show that if facing the west, and holding the paper in a north and south vertical plane, then ei , e2i , · · ·e−i , e−2i , turn respectively 1, 2, · · · radians counter-clockwise, and 1, 2, · · · radians clockwise in the plane of the paper. Also π show that e± 2 i = ±i, e±πi = −1 e2nπi = 1, where n is any integer. 7. Show by diagram that Seθi = cos θ, V eθi = i sin θ, where θ is any positive or negative number and the unit of angle is a radian. 8. Show that if OA rotate into OB through 2 arc q, then (OB) = q(OA)q −1 . 9. Show that if α be a vector in the plane of q, then Kq = αqα−1 = α−1 qα. 10. Show that pq rotates into qp, and determine two such rotations. 11. Show that SKq = Sq, V Kq = −V q. 12. Show that Kαβ = βα, Sαβ = Sβα, V αβ = −V βα. 13. Show that Kαβγ = −γβα; V αβγ = V γβα; Sαβγ = Sβγα = Sγαβ = −Sγβα = −Sβαγ = −Sαγβ. (a) Determine the conjugate of a product of n vectors. 14. Prove by diagram that Kpq = Kq · Kp.
Addition 47. Definition. The sum (p + q) is the number determined by the condition that its product is the sum of the products of p and q.
CHAPTER 3. QUATERNIONS
32
Thus let OA be any step that is multiplied by both p and q, and let pOA = OB, qOA = OC, and OB + OC = OD, then (p + q)OA = OD. It is obvious that any change in OA alters OB, OC, OD, proportionally, so that the value of the sum p + q(= OD : OA) is the same for all possible values of OA. Similarly, any quaternion, r, may be added to the sum p + q, giving the sum (p+q)+r; and we may form other sums such as p+(q +r), (q +r)+p, etc. It will be shown later that all such sums of the same numbers are equal, or that quaternion addition is associative and commutative. 48. The sum of a scalar and a vector is a quaternion with that scalar and that vector, and conversely, a quaternion is the sum of its scalar and its vector.
For let w be any scalar, and ρ any vector, and let wOA = OL, ρOA = OM, then completing the rectangle OLBM, we have
CHAPTER 3. QUATERNIONS
33
(w + ρ)OA = OB, and the scalar of w + ρ is w, and its vector is ρ, since OL, OM are the components of OB parallel and perpendicular to OA. Similarly, q = Sq + V q. 49. The scalar, vector, and conjugate, of any sum equals the like sum of the scalars, vectors, and conjugates of the terms of the sum. [I.e., S, V , K, are distributive over a sum.]
For let pOA = OB, qOA = OC, (p + q)OA = OB + OC = OD. Then the components of OD parallel and perpendicular to OA are, by the figure, the sums of the like components of OB, OC; i.e., S(p + q) · OA = Sp · OA + Sq · OA, or S(p + q) = Sp + Sq; and V (p + q) · OA = V p · OA + V q · OA, or V (p + q) = V p + V q. Also, if OB 0 D0 C 0 be the parallelogram that is symmetric to the parallelogram OBDC with reference to OA as axis of symmetry, then Kp · OA = OB0 , Kq · OA = OC0 , and K(p + q) · OA = OD0 , and since OB0 + OC0 = OD0 , therefore K(p + q) = Kp + Kq. These results extend to any given sum; e.g., V [(p+q)+r] = V (p+q)+V r = (V p + V q) + V r, etc.
CHAPTER 3. QUATERNIONS
34
50. If (OA) + (OB) = (OC), then OA + OB = OC, and conversely.
For erect a pin OD of unit length perpendicular to the plane of the angle AOB on its counter-clockwise side; and turn AOB round OD as axis through a clockwise right angle as seen from D into the position A0 OB 0 . Then since (OA) is the vector that turns through a counter-clockwise right angle round OA as axis, and extends unit length into OA = OA0 , therefore (OA)OD = OA0 , and similarly (OB)OD = OB 0 , and therefore (OC)OD = OA0 + OB 0 = OC 0 , where OA0 C 0 B 0 is a parallelogram. Hence the step OC of proper length and direction to give the tensor and axis of the vector (OC) must be the diagonal of the parallelogram on OA, OB as sides; and therefore OA + OB = OC. Conversely, if OA + OB = OC, then turning the parallelogram OACB into the position OA0 C 0 B 0 , we have, since OA0 + OB 0 = OC 0 , that (OA) + (OB) = (OC). • Cor. 1. Vectors add in the same way as their corresponding steps, and all the laws of addition and resolution of steps extend at once to vectors. • Cor. 2. A sum of quaternions is associative and commutative.
CHAPTER 3. QUATERNIONS
35
For since by Cor. 1 a sum of vectors is independent of the way in which its terms are added, and since we know that a sum of scalars (i.e., ordinary numbers) is independent of the way in which its terms are added, therefore by Art. 49 the scalar and the vector of a sum are independent of the way in which the sum is added. Hence the sum is independent of the way in which it is added, since it is equal to the sum of its scalar and its vector. 51. Lemma. If p, q be any quaternions, then (1 + p)q = q + pq.
For take OB in the intersection of the planes of p, q and draw OA, OC such that q · OA = OB, pOB = OC; then (1 + p)q · OA = (1 + p)OB = OB + OC = qOA + pqOA. Hence, (1 + p)q = q + pq. 52. If p, q, r be any quaternions, then (p + q)r = pr + qr. For we have, (1 + qp−1 )p · r = (1 + qp−1 ) · pr, and expanding each member by the preceding lemma, we have, (p + q)r = pr + qr. This result extends to any sum; e.g., (p + q + r + s)t = [(p + q) + (r + s)]t = (p + q)t + (r + s)t = pt + qt + rt + st. • Cor. 1. r(p + q) = rp + rq. For let p0 , q 0 , r0 be the conjugates of p, q, r. Then from (p0 + q 0 )r0 = p0 r0 + q 0 r0 , we have, by taking the conjugates of each member, r(p + q) = rp + rq. [Art. 39, Cor. 2; Art. 49.] • Cor. 2. A product of sums equals the sum of all partial products that may be formed from the given product by multiplying together, in the order in which they stand, a term from each factor of the product. E.g., (p + q)(r + s) = pr + ps + qr + qs.
CHAPTER 3. QUATERNIONS
36
Note.—This rule should be used even when the factors are commutative, as it prevents all danger of taking out the same partial product twice; e.g., from taking both pr and rp from the above product. To be sure that all the partial products are found, some system of arrangement should be adopted; also the total number of partial products should be determined.
E.g., (p + q)(p + q)(p + q) may be arranged according to the degrees of the terms in p, and there are 2 × 2 × 2 = 8 terms. This product is then easily seen to be p3 + (p2 q + pqp + qp2 ) + (pq 2 + qpq + q 2 p) + q 3 , when p, q are not commutative, and p3 + 3p2 q + 3pq 2 + q 3 , when p, q are commutative.
Formulas. For Exercise and Reference 53. (a) q = T q · U q = U q · T q. (b) q = Sq + V q; Kq = Sq − V q. (c) Sq = T q cos ∠q, = r cos θ, say; T V q = T q sin ∠q = r sin θ. (d) V q = T V q · U V q = r sin θ where = U V q. (e) q = r(cos θ + · sin θ) = reθ , Kq = r(cos θ − sin sin θ) = re−θ 0
0
(f) eθ · eθ = e(θ+θ ) . (g) Sq = 12 (q + Kq); V q = 21 (q − Kq). (h) T q 2 = qKq = Kq · q = (Sq)2 − (V q)2 = (Sq)2 + (T V q)2 . (i) q −1 = Kq/T q 2 . As a further exercise find the T , U , S, V , K of the T , U , S, V , K of q, in terms of r, θ, . 54. (a) T (· · · pqr) = · · · T p · T q · T r. (b) U (· · · pqr) = · · · U p · U q · U r. (c) ∠(· · · pqr) = ∠(r · · · pq) = ∠(qr · · · p), etc. (d) S(· · · pqr) = S(r · · · pq) = S(qr · · · p), etc. (e) T V (· · · pqr) = T V (r · · · pq) = T V (qr · · · p), etc. (f) arc (· · · pqr) = arc r + arc q + arc p + · · · . (g) (· · · pqr)−1 = r−1 q −1 p−1 · · · . (h) K(· · · pqr) = Kr · Kq · Kp · · · .
CHAPTER 3. QUATERNIONS
37
(i) S(xp + yq + zr) = xSp + ySq + zSr, [x, y, z, scalars] and similarly for V or K instead of S. 55. (a) Kα = −α; T α2 = −α2 ; Sα = 0; V α = α. (b) Kαβ = βα; Sαβ = Sβα; V αβ = −V βα. (c) αβ + βα = 2Sαβ, αβ − βα = 2V αβ. (d) (α ± β)2 = α2 ± 2Sαβ + β 2 . x y V aβ, say. [x, y, (e) V (xα + yβ)(x0 α + y 0 β) = (xy 0 − x0 y)V αβ = 0 x y0 x0 , y 0 , scalars.] z x y z V βγ + (f) V (xα + yβ + zγ)(x0 α + y 0 β + z 0 γ) = 0 z 0 x0 V γα + y z0 x y 0 0 0 0 x y 0 V αβ. [x, y, z, x , y , z , scalars.] 56. (a) Kαβγ = −γβα. (b) αβγ − γβα = 2Sαβγ = −2Sγβα. Hence the scalars of the six products of α, β, γ are equal to one of two negative numbers according to the cyclic order of the product; and an interchange in two factors (which changes the cyclic order) changes the sign of the scalar of the product. When two of the three factors are equal, the scalar of their product must therefore be zero, since an interchange of the equal factors changes the sign without changing the value. 0 0 0 00 (c) S · (xα + yβ + zγ)(x α + y 00 β + z 00 γ) 0α + 0y β + z γ)(x n y 0 z 0 0 0 o z y x x = x 00 00 Sαβγ 00 + z 00 00 + y 00 x y z x y z x 0 y0 z0 y z Sαβγ, say. [x, y, z, etc., scalars.] = x x00 y 00 z 00
(d) Sαβγ = SαV βγ = SβV γα = SγV αβ. [Replace βγ by Sβγ + V βγ, expand by 54 (i ), and note that S · αSβγ = 0.] (e) αβγ + γβα = 2V αβγ = 2V γβα. Note.—Insert between the two terms of the first member of (e), the null term (αγβ − αγβ − γαβ + γαβ), and it becomes α(βγ + γβ) − (αγ + γα)β + γ(αβ + βα). Hence, using (55 c), we have (f ).
(f) V αβγ = αSβγ − βSγα + γSαβ. Transpose the first term of the second member of (f ) to the first member, noting that αSβγ = V · αSβγ, and βγ − Sβγ = V βγ, and we have (g) V αV βγ = −βSγα + γSαβ; (g 0 ) V · (V βγ)α = βSγα − γSαβ.
CHAPTER 3. QUATERNIONS
38
(h) [(g), (d )] [(g 0 ), (d )]
V · (V αβ)V γδ = −γSαβδ + δSαβγ = αSβγδ − βSαγδ.
(i) [(h)]
δSαβγ = αSβγδ + βSγαδ + γSαβδ.
Replace α, β, γ, by V βγ, V γα, V γβ, noting that V · (V γα · V αβ) = −αSαβγ, etc., and that S(V βγ · V γα · V αβ) = −(Sαβγ)2 , and we have (j) δSαβγ = V βγSαδ + V γαSβδ + V αβSγδ. Note.—(i), (j ) may be obtained directly by putting δ = xα + yβ + zγ or xV βγ + yV γα + zV αβ, and finding x, y, z, by multiplying in the first case by βγ, γα, αβ, and in the second case by α, β, γ, and taking the scalars of the several products.
57. (a) i2 = j2 = k2 = ijk = −1; jk = i = −kj; ki = j = −ik, ij = k = −ji. (b) ρ = iSiρ − jSjρ − kSkρ. [56 (i ) or (j ) or directly as in note.] Let ρ = xi + yj + zk, ρ0 = x0 i + y 0 j + z 0 k, etc. [x, y, z, etc. scalars.] Then, prove by direct multiplication, (c) −ρ2 = x2 + y 2 + z 2 = T ρ2 . (d) −Sρρ0 = xx0 + yy 0 + zz 0 = −sρ0 ρ. x y z x y z 0 j + i + (e) vρρ0 = 0 x0 y 0 k = −V ρ ρ. z 0 x0 y z0 x y z (f) −Sρρ0 ρ00 = x0 y 0 z 0 = −SρV ρ0 ρ00 . x00 y 00 z 00
Geometric Theorems 58. The angle of αβ equals the supplement of the angle θ between α, β.
CHAPTER 3. QUATERNIONS
39
For, since αβ · β −1 = α, therefore αβ turns through the angle from β −1 to α, which is the supplement of the angle θ from α to β. • Cor. Sαβ = −T αβ cos θ, T V αβ = T αβ sin θ. [Sq = T q cos ∠q, etc.] 59. The scalar of αβ equals the product of α and the projection of β upon it; the vector of αβ equals the product of α and the projection of β perpendicular to it, and V αβ is a vector perpendicular to α, β on their counter-clockwise side whose length equals the area of the parallelogram on α, β as sides.
CHAPTER 3. QUATERNIONS
40
Let β1 , β2 be the components of β parallel and perpendicular to α, then β = β1 + β2 and αβ = αβ1 + αβ2 = scalar + vector . Hence Sαβ = αβ1 , as stated; and V αβ = αβ2 , which is β2 turned a counter-clockwise right angle round α and lengthened by T α. Hence V αβ is perpendicular to α, β on their counterclockwise side (towards the reader in the figure), and its length is T α · T β2 = area parallelogram on α, β, as sides.1 • Cor. 1. The projections of β parallel and perpendicular to α equal α−1 Sαβ and α−1 V αβ. • Cor. 2. The scalar measure of the projection of β upon α is −Sαβ Tα , and the tensor measure of the projection of β perpendicular to α is T V αβ T α . [Also from 58, Cor.] • Cor. 3. If θ be the angle between α, β, then cos θ = − TSαβ αβ , sin θ = T V αβ T αβ .
[Also from 58, Cor.]
60. The volume of a parallelepiped on α, β, γ as edges is −Sαβγ (the volume being positive or negative according as α lies on the counter-clockwise or clockwise side of β, γ).
For let α1 be the projection of α upon V βγ; then taking the face β, γ of the parallelepiped as the base, we have by Art. 59 that T V βγ is the area 1 The parallelogram on α, β may be considered as bounded by the path of a point that receives the displacement α, then the displacement β, then the displacement −α, then the displacement −β. This area is therefore bounded counter-clockwise round V αβ as axis; and V αβ may therefore be called the vector measure of the area of this directed parallelogram or of any parallel plane area of the same magnitude and direction of boundary.
CHAPTER 3. QUATERNIONS
41
of the base; also T α1 is the altitude. Hence numerical volume [Art. 46.] [59, 56, d.]
= T α1 · T V βγ = ∓α1 V βγ = ∓SαV βγ = ∓Sαβγ.
The upper or lower sign must be taken according as α1 , V βγ are in the same or opposite directions. This numerical result must be multiplied by 1 when α lies on the clockwise side of β, γ; i.e., when α1 , V βγ are opposites (since V βγ lies on the counter-clockwise side of β, γ). Hence—Sαβγ is the required algebraic volume. • Cor. The condition that α, β, γ are coplanar vectors is that Sαβγ = 0 (or αβγ = a vector).
Examples 1. Expand (p + q + r)2 , (α + β + γ)2 , (p + q)(p − q), (p − q)(p + q), (p + q)K(p + q). Show that T (p + q)2 = T p2 + 2SpKq + T q 2 . 2. Solve q 2 + 4kq − 8 = 0 for q. [q must be cocircular with k. Hence q = −2k ± 2 are the real solutions.] 3. Find the tensor, versor, scalar, vector, and angle of each of the numbers: 2, −3, π 3i, 2 + 3i, i + j, 3i + 4j, 5e 3 i , (2i + 3j + 6k)2 . 4. Show that the√three quaternion cube roots of −1, with horizontal great circle, are −1, 21 ± 12 3k. 5. Show geometrically that eθ + e−θ = 2 cos θ, θ − e−θ = 2 sin θ. 6. The numbers eα and 1+α+
α2 α3 α4 + + + ··· 2! 3! 4!
are equal. Verify this approximately by geometric construction when T α = 1, and when T α = 2. [For the series, construct OA, AB = αOA, BC = 12 αAB, CD = 13 αBC, DE = 41 αCD, etc.] 7. In the plane triangle ABC, whose sides opposite A, B, C, are a, b, c, show by squaring BC = AC − AB, that as a2 = b2 + c2 − 2bc cos A; also from V BCCA = V CAAB = V ABBC show that a : b : c = sin A : sin B : sin C. 8. From eθ = cos θ + sin θ, eθ · eθ
0
= e(θ+θ
0
)
, show that
cos(θ + θ0 ) = cos θ cos θ0 − sin θ sin θ0 , sin(θ + θ0 ) = sin θ cos θ0 + cos θ sin θ0 . 9. Show that (cos θ + sin θ)n = cos nθ + sin nθ. 10. Show that Spq = SpSq + S(V p · V q), and hence that cos ∠pq = cos ∠p · cos ∠q + sin ∠p · sin ∠q · cos ∠(V p · V q).
CHAPTER 3. QUATERNIONS _
42
_
11. If arc q = BA, arc p = AC, show that the last equation of Ex. 10 is the property cos a = cos b cos c + sin b sin c cos A _
_
of the spherical triangle ABC. [Draw B 0 A = arc V q, AC 0 = arc V p.] 12. If α, β, γ, α0 , β 0 , γ 0 are vectors from O to the vertices A, B, C, A0 , B 0 , C 0 of two spherical triangles on the unit sphere O, where α0 = U V βγ, β 0 = U V γα, γ 0 = U V (αβ); then α = U V β 0 γ 0 , β = U V γ 0 α0 , γ = U V α0 β 0 , and the two triangles are polar triangles. 13. In Ex. 12 show that cos α = −Sβγ, sin α = T V βγ, etc.; cos A = S(U V γα · U V αβ) = Sβ 0 γ 0 = − cos α0 , etc. Hence ∠A, ∠α0 are supplements, etc. 14. Show that the equation of Ex. 11 follows from the identity, −Sβγ = S(βγ ·γα) = Sβγ · Sγα + S(V γα · V αβ). 15. From V (V γα · V αβ) = −αSαβγ, and the similar equations found by advancing the cyclic order α, β, γ, show that we have in the spherical triangle ABC, sin a : sin b : sin c = sin A : sin B : sin C. 16. Show that if α, β, γ are coplanar unit vectors, then αβγ = −αβ −1 · γ = (γ turned through the angle from β to α and reversed) = (β rotated 180◦ about the exterior bisector of the angle between α, γ) = (α − γ)β(α − γ)−1 . 17. Show that (V ABCD)−1 SACV ABCD is the shortest vector from the line AB to the line CD. [Project AC upon the common perpendicular to AB, CD.] 18. If α, β, γ be the vector edges about a vertex of an equilateral pyramid (whose edges are unit lengths), then β − γ, γ − α, α − β, are the remaining vector edges. Hence show that Sβγ = Sγα = Sαβ = − 21 , and V αV βγ = (−β)Sγα + γSαβ = 1 (β − γ). Also show that: 2 √ √ (a) The face angles are 60◦ , the area of a face is 14 3, and its altitude is 12 3. √ (b) Opposite edges are perpendicular, and their shortest distance is 21 2. √ (c) The angle between a face and an edge is cos−1 31 3. √ (d) The angle between two adjacent faces is sin−1 23 2. √ 1√ 1 (e) The volume and altitude of the pyramid are 12 2, 3 3. 19. The cosines of the angles that a vector makes with i, j, k, are called its direction cosines. Find the lengths and direction cosines of 2i − 3j + 6k, i + 2j − 2k, xi + yj + zk. 20. Show that the sum of the squares of the direction cosines of a line equals 1. 21. If (l, m, n), (l0 , m0 , n0 ) are the direction cosines of two lines, show that li+mj+nk, l0 i + m0 j + n0 k, are unit vectors in the directions of the lines, and that if θ be the angle cos θ = ll0 + mm0 + nn0 ; also that sin2 θ = between the lines, then m n 2 n l 2 l m 2 0 m n0 + n0 l0 + l0 m0 , and that the three terms of the second member, respectively divided by sin2 θ, are the squares of the direction cosines of a line that is perpendicular to the given lines. [Art. 57.]
CHAPTER 3. QUATERNIONS
43
22. If O be a given origin, then the vector OP = ρ = xi + yj + zk say, is called the vector of P with reference to the given origin. If OX, OY , OZ be axes in the directions of i, j, k, the scalar values of the projections of OP upon these axes, i.e., (x, y, z), are called the co¨ ordinates of P with reference to the given axes. Let the co¨ ordinates of the vertices of the pyramid ABCD be, respectively, (8,2,7), (10,6,3), (1,6,3), (9,10,11). Draw this pyramid with reference to a perspective of i, j, k, showing co¨ ordinates and vectors. Also: (a) Find the vectors and co¨ ordinates of the middle points of the edges. [OM = 21 (OA + OB), etc.] (b) Find the lengths and direction cosines of the edges. [−AB2 = AB 2 , etc.] (c) Find vectors that bisect the face angles. [U AC±U AD bisects ∠CAD.] (d) Find altitudes of the faces and the vectors of their feet. [If L be the foot of the perpendicular from B on AC, then AL = AB−1 SABAC, etc.] (e) Find the areas of the faces. (f) Find the volume and altitudes of the pyramid. (g) Find the angles between opposite edges, and their (shortest) distance apart. [Ex. 17.] (h) Find the angle between two adjacent faces.
Chapter 4
Equations of First Degree 61. The general equation of first degree in an unknown vector ρ is of the form, (a) q1 ρr1 + q2 ρr2 + · · · = q, where q, q1 , r1 , q2 , r2 , · · · are known numbers. This equation may be resolved into two equations by taking the scalar and the vector of each member; and we shall consider these equations separately. 62. Taking the scalar of (a), Art. 61, the term Sq1 ρr1 becomes, by a cyclic change in the factors, S · r1 q1 ρ, and this becomes [by dropping the vector (Sr1 q1 )ρ, since its scalar is zero] S(V r1 q1 · ρ); and similarly for the other terms. Hence if we put V r1 q1 + V r2 q2 + · · · = δ, and Sq = d, the general scalar equation of first degree in ρ becomes, (a) Sδρ = d or Sδ(ρ − dδ −1 ) = 0. One solution of this equation is obviously ρ = dδ −1 . This is not the only solution, since by Art. 45, Cor. 3, the second factor may be any vector that is perpendicular to δ. Hence the general solution is ρ = dδ −1 + V σδ, where σ is an arbitrary vector. 63. Hence, draw OD = δ, take N on the line OD so that ON = dδ −1 , and draw any vector NP = V σδ that is perpendicular to the line OD; then ρ = OP is a solution of the equation Sδρ = d.
44
CHAPTER 4. EQUATIONS OF FIRST DEGREE
45
The locus of P is therefore a plane perpendicular to OD at the point N ; and this plane is called the locus of the equation Sδρ = d, with respect to the origin O. [The locus is the assemblage of all points that satisfy the equation.] 64. The vector perpendicular distance from the plane Sδρ − d = 0 to the point P 0 (whose vector is ρ0 ) is δ −1 (Sδρ0 − d), and the corresponding scalar distance measured upon δ is −(Sδρ0 − d) . Tδ For the perpendicular distance of P 0 is the projection of NP0 = (ρ0 −dδ −1 ), upon OD = δ. 65. The locus of the simultaneous equations Saρ = a, Sβρ = b is a straight line, viz., the intersection of the two plane loci of these equations taken separately. For in order that ρ = OP may satisfy both equations, P must lie in both planes, and its locus is therefore the intersection, of those planes. 66. The equation V δρ = δ 0 , or V δ(ρ − δ −1 δ 0 ) = 0, is a consistent equation only when δ 0 is perpendicular to δ, since V δρ is always perpendicular to δ. When δ 0 is perpendicular to δ, then δ −1 δ 0 is a vector (Art. 45, Cor. 1), and the general solution of this equation is ρ = δ −1 δ 0 + xδ, where x is an arbitrary scalar (Art. 46, Cor.). Hence draw ON = δ −1 δ 0 , and NP = xδ (any vector parallel to δ), and then ρ = OP is a solution of the given equation. The locus of P is therefore the straight line through N parallel to δ, and ON is the perpendicular from the origin
CHAPTER 4. EQUATIONS OF FIRST DEGREE
46
upon the line. The equations of Art. 65 take this form by multiplying the first by β, the second by α, and subtracting, remembering that V (V αβ · ρ) = αSβρ − βSαρ.
67. The vector perpendicular distance from the line V δρ − δ 0 = 0 to the point P0 is δ −1 (V δρ0 − δ 0 ), where ρ0 = OP0 . For the required perpendicular distance of P 0 is the projection of NP0 , = (ρ0 − δ −1 δ 0 ), perpendicular to δ. 68. The point of intersection of the three planes Sαρ = a, Sβρ = b, Sγρ = c is [Art. 56, (j ).]
ρ=
(aV βγ + bV γα + cV αβ) . Sαβγ
Examples 1. Find the equation of the locus of a point that moves so that its numerical distances from two fixed points are equal. 2. A point moves so that its scalar distances from two fixed planes are equal; show that its locus is a plane bisector of the diedral angle of the given planes. 3. A point moves so that the sum or difference of its scalar distances from two fixed planes is constant; show that its locus is a plane parallel to the interior or exterior bisector of the diedral angle of the given planes. 4. A point moves so that the ratio of its scalar distances from two fixed planes is constant; show that its locus is a plane. 5. A point moves so that its numerical distances from two intersecting lines are equal; find its locus. [Take the point of intersection as origin.] 6. A point moves so that its numerical distances from three fixed points are equal; find its locus. (a) The same with coplanar lines instead of points. [Four straight lines perpendicular to the plane of the lines.]
CHAPTER 4. EQUATIONS OF FIRST DEGREE
47
7. Find the vector of the centre of the sphere whose surface passes through four given points. 8. A point moves so that its tangential distances from two given spheres are numerically equal; find its locus. 9. On the chord OQ of a given sphere a point P is taken so that OP · OQ = −a2 ; when Q moves round the sphere find the locus of P . [A plane perpendicular to the diameter OD.] 10. The locus of the point P whose co¨ ordinates (x, y, z) satisfy lx+my+nz+d = 0 is a plane √ perpendicular to the vector li + mj + nk, at a distance from the origin of −d/ l2 + m2 + n2 , measured in the direction of this vector. (a) Show that the equation of this plane may be put in the form x cos α + y cos β + z cos γ − p = 0, where p is the perpendicular distance from O to the plane and the cosines are the direction cosines of this perpendicular. 11. Find the perpendicular distance of P 0 = (x0 , y 0 , z 0 ), from the plane of Ex. 10, [x0 cos α + y 0 cos β + z 0 cos γ − p]. 12. The locus of the point P whose co¨ ordinates (x, y, z) satisfy (x−a) = (y−b) = l m (z−c) is a line parallel to the vector li + mj + nk through the point (a, b, c). If n P satisfy the first two of these three equations, its locus is a plane through the line, perpendicular to the plane of XOY . [If t be the common value of the three ratios, then ρ = ai + bj + ck + t(li + mj + nk).] 13. Find the perpendicular distance of P 0 = (x0 , y 0 , z 0 ) from the line of Ex. 12. In the following examples A, B, C, D, P are points whose co¨ ordinates are (8, 2, 7), (10, 6, 3), (1, 6, 3), (9, 10, 11), (x, y, z). 14. The equation of the plane through A perpendicular to OD is SODAP = 0, or 9x + 10y + 11z = 169. 15. The equation of the plane through AB parallel to CD is S · APV ABCD = 0, or 2x − 2y − z = 5. 16. The equation of the plane ABC is S · APV ABAC = 0 or y + z = 9. 17. Find the perpendicular distance of D from the planes in Exs. 14, 15, 16. 18. The equation of the plane through AB that contains the common perpendicular to AB, CD is S · APV (ABV ABCD) = 0, or 2x + y + 2z = 32. 19. The equation of the line through A parallel to OD is V ODAP = 0 or AP = = (y−2) = (z−7) . tOD, or (x−8) 9 10 11 20. The equation of the line AB is V ABAP = 0, or AP = tAB or (z−7) . −4
(x−8) 2
=
(y−2) 4
=
21. The equation of the common perpendicular to AB, CD is the equation of Ex. 18 and x + 2y − 2z = 7. 22. Find the distance of D from the lines in Exs. 19, 20, 21. 23. Find OD in the form lOA + mOB + nOC, and find the ratios in which OD cuts the triangle ABC.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
48
Nonions 69. The vector equation of first degree is (a)
V q1 ρr1 + V q2 ρr2 + · · · = V q.
To solve this equation we resolve it along i, j, k, by multiplying it by these vectors and taking the scalars of the products. We thus find three scalar equations of first degree from which ρ may be immediately found as in Art. 68. Hence (a) has in general one, and only one, solution which corresponds to the intersection of three given planes. [See further Art. 81.] 70. The first member of Art. 69 (a) is a linear, homogeneous vector function of ρ; i.e., it is of first degree in ρ, every term is of the same degree in ρ, and it is a vector. We may denote the operator V q1 ()r1 + V q2 ()r2 + · · · by a single letter, φ, so that φρ, φσ, · · · denote the vectors that result from putting ρ, σ, · · · in the places occupied by the parenthesis. 71. The operator φ is distributive over a sum and commutative with scalars; i.e., φ(xρ + yσ) = xφρ + yφσ. This is immediately verified by putting xρ + yσ in the places occupied by the parentheses of φ and expanding the several terms. 72. We have ρ = xα + yβ + zγ, where α, β, γ are given non-coplanar vectors, and x, y, z, are scalars, each of first degree in ρ, as shown in 56(i ) with ρ in the place of δ; hence, (a)
φρ = xφα + yφβ + zφγ.
The complete operation of φ is therefore determined when the three vectors φα, φβ, φγ are known. Since each of these vectors involves three scalar constants (e.g., the multiples of the given non-coplanar vectors α, β, γ, that express it), therefore the value of φ depends upon nine scalar constants. The operator φ may therefore be called a nonion. Scalars and rotators are particular forms of nonions. Note.—It is readily shown that nonions have the same laws of addition and multiplication among themselves as quaternions. Products are not in general commutative. A product (φ − g1 )(φ − g2 )(φ − g3 ), where g1 , g2 , g3 are scalars, is commutative, since φ is commutative with scalars by Art. 71. Hence this product multiplies out as if φ were a scalar, and is φ3 − (g1 + g2 + g3 )φ2 + (g2 g3 + g3 g1 + g1 g2 )φ − g1 g2 g3 .
CHAPTER 4. EQUATIONS OF FIRST DEGREE
49
Linear Homogeneous Strain 73. An elastic solid is subjected to the strain φ with respect to an origin O, when all its particles, A, B, C, etc., are displaced to positions A0 , B 0 , C 0 , etc., that are determined by OA0 = φOA, OB0 = φOB, OC0 = φOC, etc. In general, any particle P whose vector is OP = ρ occupies after the strain the position P 0 , whose vector is OP0 = φρ. The particle at O is not moved, since its vector after strain is φOO = φ0 = 0.
(a) We have, also, φAP = A0 P0 , etc. For, A0 P0 = OP0 − OA0 = φOP − φOA = φ(OP − OA) = φAP, etc. 74. A straight line of particles parallel to α is homogeneously stretched and turned by the strain φ into a straight line of particles parallel to φα, and the ratio of extension and turning is φα/α. For let AP be a line parallel to α, and let A, P strain into A0 , P 0 . Then, since AP = xα, therefore, by 73a, A0 P0 = xφα, and the ratio of extension and turning is A0 P0 /AP = φα/α. Note.—This property that parallel lengths of the substance strain into parallel lengths and are stretched proportionally, is the physical definition of linear homogeneous strain.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
50
75. A plane of particles parallel to α, β is homogeneously spread and turned by the strain φ into a plane of particles parallel to φα, φβ, and the ratio of extension and turning is V φαφβ/V αβ. For let AP Q be a plane parallel to α, β, and let A, P , Q strain into A0 , P 0 , Q0 . Then, since AP = xα + yβ, AQ = x0 α + y 0 β, therefore A0 P0 = xφα + yφβ,
A0 Q0 = x0 φα + y 0 φβ.
By Arts. 59, 55, (e), the directed area of the triangle AP Q is 21 V · AP · AQ = 12 (xy 0 − x0 y)V αβ, and the directed area of the triangle A0 P 0 Q0 is the same multiple of V φαφβ. Hence the ratio of the extension and turning of directed area is V φαφβ/V αβ. 76. A volume of particles is homogeneously dilated by the strain φ in the ratio Sφαφβφγ/Sαβγ, where α, β, γ are any given non-coplanar vectors. For let the pyramid AP QR strain into the pyramid A0 P 0 Q0 R0 . Then since AP = xα + yβ + zγ,
AQ = x0 α + y 0 β + z 0 γ,
AR = x00 α + y 00 β + z 00 γ, therefore A0 P0 , A0 Q0 , A0 R0 have these values with φα, φβ, φγ instead of α, β, γ. The volume of the pyramid AP QR relative to the order AP , AQ, AR of its edges is, by Arts. 60, 56, (c), x y z 1 1 0 y 0 z 0 Sαβγ, − SAPAQAR = − x 6 6 00 x y 00 z 00 while that of the strained pyramid is the same multiple of Sφαφβφγ. Hence the ratio of dilation of volume is Sφαφβφγ/Sαβγ. 77. The ratio of dilation of φ is called its modulus. (a) It is obvious from the signification of the modulus that the modulus of a product of nonions equals the product of the moduli of the factors; e.g., mod φψ = mod φ · mod ψ. When mod φ is positive, the parts of the volume are in the same order before and after strain. When mod φ is negative, the order of the parts is reversed by the strain; i.e., if AP lie on the counter-clockwise side of the plane AQR, then A0 P 0 lies on the clockwise side of A0 Q0 R0 , so that the particles along AP have been strained through the particles of the plane AQR. Such a strain is obviously not a physical possibility.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
51
Finite and Null Strains 78. If an elastic solid which fills all space be subjected to a strain φ, the strained solid fills all space if mod φ be finite, and it fills only an indefinite plane or line through the origin or reduces to the origin if mod φ be zero. For if Sφαφβφγ be finite, then φα, φβ, φγ are non-coplanar vectors, so that φρ(= xφα + yφβ + zφγ) may be made any vector by properly choosing ρ(= xα + yβ + zγ). But if Sφαφβφγ = 0, then φα, φβ, φγ are coplanar vectors or colinear vectors or each zero, so that φρ will be a vector in a given plane or line through O or the vector of O, whatever value be given to ρ. When mod φ is zero, φ is called a null nonion; and it is called singly or doubly or triply null, according as it strains a solid into a plane or a line or a point. If φα = 0, then α is called a null direction of φ. 79. Null strains, and only null strains, can have null directions; a singly null strain has only one null direction; a doubly null strain has a plane of null directions only; a triply null strain has all directions null. For when mod φ = 0, then φα, φβ, φγ are coplanar or colinear vectors, and we have a relation lφα + mφβ + nφγ = 0, i.e., lα + mβ + nγ is a null direction of φ. Conversely, if φ have a null direction, take one of the three non-coplanar vectors α, β, γ, in that direction, say α, and we have Sφαφβφγ = 0, since φα = 0, and therefore modφ = 0. Also, if φ have only one null direction, α, then φβ, φγ, are not parallel, since φβ = lφγ makes β−lγ a second null direction. Since ρ = xα+yβ+zγ, therefore φρ = yφβ + zφγ, which is any vector in the plane through O parallel to φβ, φγ; hence φ is singly null. But if φ have two null directions, α, β, then φρ = zφγ, which is any vector in the line through O parallel to φγ, and therefore φ is doubly null. Also, since φ(xα + yβ) = 0, therefore any direction in the plane of α, β is a null direction of φ. If φ have three non-coplanar null directions α, β, γ, then φρ = 0 for all values of ρ; i.e., a triply null nonion is identically zero. 80. A singly null nonion strains each line in its null direction into a definite point of its plane; and a doubly null nonion strains each plane that is parallel to its null plane into a definite point of its line. For when φ is singly null, say φα = 0, then xφβ + yφγ is the vector of any point in the plane of φ, and all particles that strain into this point have the vectors ρ = xα + yβ + zγ, where x is arbitrary, since φα = 0; i.e., they are particles of a line parallel to α. So, if φ is doubly null, say
CHAPTER 4. EQUATIONS OF FIRST DEGREE
52
φα = 0, φβ = 0, then any point of the line of φ is zφγ, and the particles that strain into this point have the vectors ρ = xα + yβ + zγ, in which x, y are arbitrary; i.e., they are particles of a plane parallel to α, β. Note.—It follows similarly that the strain φ alters the dimensions of a line, plane, or volume by as many dimensions as the substance strained contains independent null directions of φ, and no more. Hence, a product φψ has the null directions of the first factor, and the null directions of the second factor that lie in the figure into which the first factor strains, and so on; the order of nullity of a product cannot exceed the sum of the orders of its factors, and may be less; etc.
Solution of φρ = δ 81. The solutions of φρ = δ are, by definition of the strain φ, the vectors of the particles that strain into the position whose vector is δ. Hence: (a) When φ is finite, there is one, and only one, solution. (b) When φ is singly null, and δ does not lie in the plane of φ, there is no finite solution. Divide the equation by T ρ, and make T ρ infinite, and we find φU ρ = 0; i.e., the vector of the point at infinity in the null direction of φ is a solution. (c) When φ is singly null, and δ lies in the plane of φ, there are an infinite number of solutions, viz., the vectors of the particles of a line that is parallel to the null direction of φ. (d) When φ is doubly null, and δ does not lie in the line of φ, there is no finite solution. As in (2) the vectors of the points of the line at infinity in the null plane of φ are solutions. (e) When φ is doubly null, and δ lies in the line of φ, there are an infinite number of solutions, viz., the vectors of the particles of a plane that is parallel to the null plane of φ. These results correspond to the intersections of three planes, viz.: (a) The three planes meet in a point. (b) The three planes parallel to a line. (c) The three planes meet in a common line. (d) The three planes parallel. (e) The three planes coincide.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
53
Derived Moduli of φ 82. The ratio in which the nonion φ + g dilates volume is, mod (φ + g) = S(φα + gα)(φβ + gβ)(φγ + gγ)/Sαβγ. This is independent of the values of the non-coplanar vectors α, β, γ in terms of which it is expressed. If g is a scalar, this modulus is an ordinary cubic in g, whose coefficients will therefore depend only upon φ. The constant term is mod φ, and the coefficients of g, g 2 , are called mod1 φ, mod2 φ, so that, (a)
mod (φ + g) = g 3 + g 2 mod2 φ + gmod1 φ + mod φ. [mod1 φ = S(αφβφγ + βφγφα + γφαφβ)/Sαβγ; mod2 φ = S(βγφα + γαφβ + αβφγ)/Sαβγ].
83. The roots g1 , g2 , g3 , of the cubic mod (φ − g) = 0 are called the latent roots of φ. We have from 82 (a) with −g in the place of g, and the theory of equations, mod φ = g1 g2 g3 , mod1 φ = g2 g3 + g3 g1 + g1 g2 , mod2 φ = g1 + g2 + g3 . (a) The latent roots of φ − g1 are those of φ diminished by g1 . For the roots of mod (φ − g1 − g) = 0, are g = 0, g2 − g1 , g3 − g1 . E.g., g = g2 − g1 gives mod [φ − g1 − (g2 − g1 )] = mod (φ − g2 ) = 0. (b) The order of nullity of φ cannot exceed the number of its zero latent roots. For if φ has one null direction α, then φα = 0 makes mod φ = 0, so that at least one of the latent roots is zero, say g1 ; and if φ has a second null direction β, then φα = 0, φβ = 0, makes mod1 φ = 0 or g2 g3 = 0, so that another latent root is zero, etc. (c) The order of nullity of φ−g1 cannot exceed the number of latent roots of φ that equal g1 . [(a), (b)]
CHAPTER 4. EQUATIONS OF FIRST DEGREE
54
Latent Lines and Planes of φ 84. Those lines and planes that remain unaltered in geometrical position by the strain φ are called latent lines and planes of φ. (a) The latent directions of φ are the null directions of φ − g1 , φ − g2 , φ − g3 , and g1 , g2 , g3 are the corresponding ratios of extension in those directions. For if ρ is any latent direction, and g is the ratio of extension in that direction, then we have φρ = gρ or (φ − g)ρ = 0. Hence φ − g is a null nonion, or mod (φ − g) = 0, so that g is a latent root of φ; also ρ is a null direction of φ − g. Note.—Since a cubic with real coefficients has at least one real root, therefore a real nonion has at least one latent direction. Also if two roots are imaginary, they are conjugate imaginaries, and the corresponding latent directions must also be conjugate imaginaries.
85. If α, β, γ be the latent directions corresponding to g1 , g2 , g3 , then (β, γ), (γ, α), (α, β) determine latent planes of φ in which the ratios of spreading are g2 g3 , g3 g1 , g1 g2 . E.g., V φβφγ = V (g2 β · g3 γ) = g2 g3 V βγ. Hence, in the general case when the latent roots are all unequal, the latent vectors α, β, γ must form a non-coplanar system, since any two of the latent lines or planes determined by them have unequal ratios of extension, and cannot, therefore, coincide. (a) The plane of φ − g1 is the latent plane corresponding to g2 , g3 . For (φ − g1 )ρ = y(g2 − g1 )β + z(g3 − g1 )γ, (= plane of β, γ). [The plane and null line of φ − g1 may be called corresponding latents of φ.]
The Characteristic Equation of φ 86. We have also, (a) (φ − g1 )(φ − g2 )(φ − g3 ) = 0. For the first member has the three non-coplanar null directions α, β, γ. [See 80 note, 72 note.]
CHAPTER 4. EQUATIONS OF FIRST DEGREE
55
Conjugate Nonions 87. Two nonions φ, φ0 are conjugate when (a) Sρφσ = Sσφ0 ρ for all values of ρ, σ. When φ is known, this determines φ0 without ambiguity. Thus, put σ = i, j, k, in turn, and we have by Art. 57 (b), φ0 ρ = −iSρφi − jSρφj − kSρφk. Conversely, this function satisfies (a), for we have Sσφ0 ρ = Sρφ(−iSiσ − jSjσ − kSkσ) = Sρφσ. 88. From this definition of conjugate strains we have (a) (aφ + bψ)0 = aφ0 + bψ 0 ; (φψ)0 = ψ 0 φ0 . (b) (V q()q)0 = V q()p, [αSβ()]0 = βSα(). E.g., Sσ(φψ)0 ρ = Sρφψσ = Sρφ(ψσ) = Sψσφ0 ρ = Sσψ 0 φ0 ρ, and therefore (φψ)0 = ψ 0 φ0 . [If Sσ(α − β) = 0 for all values of σ, then α − β = 0, since no vector is perpendicular to every vector σ. Hence, comparing the first and last member of the above equation, we have (φψ)0 ρ = ψ 0 φ0 ρ.] 89. Two conjugate strains have the same latent roots and moduli, and a latent plane of one is perpendicular to the corresponding latent line of the other. For since (φ − g1 )α = 0, therefore 0 = Sρ(φ − g1 )α = Sα(φ0 − g1 )ρ, and therefore φ0 − g1 is a null nonion whose plane is perpendicular to α. Hence g1 is a latent root of φ0 , and the latent plane of φ0 corresponding to φ0 − g1 is perpendicular to the latent line of φ corresponding to φ − g1 . [Art. 85, (a).]
Self-conjugate Nonions 90. A nonion φ is self-conjugate when φ0 = φ or when Sρφσ = Sσφρ for all values of ρ, σ. In consequence of this relation a self-conjugate strain has only six scalar constants, three of the nine being equal to three others, viz., Siφj = Sjφi, Siφk = Skφi, Sjφk = Skφj.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
56
91. A self-conjugate strain has by Art. 88 three mutually perpendicular latent directions, and conversely, if φ have three mutually perpendicular latent directions, i, j, k, corresponding to latent roots a, b, c, then φρ = −aiSiρ − bjSjρ − ckSkρ, which is self-conjugate. [68 b.] 92. A real self-conjugate strain has real latent roots. √ √ For let α0 = α + β −1, β 0 = α − β −1 be latent directions corresponding to conjugate imaginary roots a, b of a real nonion φ; then, if φ is selfconjugate, we have Sα0 φβ 0 = Sβ 0 φα0 = bSα0 β 0 = aSα0 β 0 , or, since a, b are unequal, therefore Sα0 β 0 = 0; but this is impossible, since Sα0 β 0 = α2 + β 2 , a negative quantity. Therefore φ is not self-conjugate if it has imaginary latent roots. 93. A nonion φ is negatively self-conjugate when φ0 = −φ, or when Sσφρ = −Sρφσ. Such a nonion has therefore only three scalar constants, since Siφi = −Siφi shows that Siφi = 0, and similarly, Sjφj = 0, Skφk = 0, while the other six constants occur in negative pairs Siφj = −Sjφi, etc. (a) The identity Sρφρ = 0 gives (by putting ρ = xi + yj + zk where x, y, z are arbitrary) all the above relations between the constants of φ, and is therefore the sufficient condition that φ is negatively selfconjugate. It shows that φρ is perpendicular to ρ or that φρ = V ρ, where must be independent of ρ since φρ is linear in ρ. 94. Any nonion φ may be resolved into a sum of a conjugate and a negatively self-conjugate nonion in only one way. For if φ = φ + ψ, where φ0 = φ, ψ 0 = −ψ, then φ0 = φ − ψ, and adding and subtracting, we have φ = 12 (φ + φ0 )ψ = 12 (φ − φ0 ), and (a)
φρ =
1 1 (φ + φ0 )ρ + (φ − φ0 )ρ = φρ + V ρ. 2 2
To find in terms of the constants of φ, we have ρ = −iSiρ − jSjρ − kSkρ, and therefore
[88 b.]
φρ = −φiSiρ − etc. φ0 ρ = −iSρφi − etc.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
57
Hence 1 1 (φ − φ0 )ρ = (iSρφi − φiSiρ) + etc. 2 2 1 = V · (V iφi)ρ + etc. = V ρ, 2 and therefore (a)
=
1 V (iφi + jφj + kφk). 2
Examples 1. Find the equation of a sphere whose centre is A(OA = α) and radius α. 2. Show that the square of the vector tangent from the sphere of Ex. 1 to P 0 is (ρ0 − α)2 + a2 . 3. Find the locus of the point P such that P P 0 is cut in opposite ratios by the sphere of Ex. 1; show that it is the plane of contact of the tangent cone from P 0 to the sphere and is perpendicular to AP 0 . 4. Let P 0 be any point on the sphere A of Ex. 1, and take P on OP 0 so that OP · OP0 + c2 = 0; find the locus of P . [P , P 0 are called inverse points with respect to O, and the locus of P is the inverse of the given sphere A. It is a sphere with centre A0 on OA, or a plane perpendicular to OA if the given sphere A pass through O.] 5. Show that the inverse of a plane is a sphere through O. 6. Show that the general scalar equation of second degree is Sρφρ + 2Sδρ + d = 0, where φ is a self-conjugate nonion. 7. Show that Sρφρ = 0 is the equation of a cone with vertex at O. 8. Show that the line ρ = α + xβ cuts the quadric surface of Ex. 6 in two points; apply the theory of equations to determine the condition that this line is a tangent to the surface, or an element of the surface, or that it meets the surface in one finite point and one point at infinity, or that the point whose vector is α lies midway between the points of intersection. 9. Show that the solution of φρ + δ = 0 is the vector of a centre of symmetry of the quadric surface of Ex. 6. Hence classify quadric surfaces as central, non-central, axial, non-axial, centro-planar. 10. Show that the locus of the middle points of chords parallel to β is a diametric plane perpendicular to φβ. 11. Show that an axial quadric is a cylinder with elements parallel to the null direction of its nonion φ. 12. Show that a non-axial quadric is a cylinder with elements parallel to the null plane of its nonion φ and perpendicular to its vector δ. 13. Show that a centro-planar quadric consists of two planes parallel to the null plane of its nonion φ.
CHAPTER 4. EQUATIONS OF FIRST DEGREE
58
14. Show that the equation of a central quadric referred to its centre as origin is Sρφρ + 1 = 0. Show that the latent lines and planes of φ are axes and planes of symmetry of the quadric; also that φρ is perpendicular to the tangent plane at the point whose vector is ρ. (a) Show that the axes and planes of symmetry of the general quadric are parallel to the latent lines and planes of φ. 15. Show that if ψ 2 = φ, then the equation of the central quadric is (ψρ)2 + 1 = 0; and that therefore the quadric surface when strained by ψ becomes a spherical surface of unit radius. 16. Show that if g, α are corresponding latent root and direction of φ, then g n , α are the same for φn . Find the latent lines and planes, the latent roots and moduli of the following nonions and their powers: (a) (aαSβγρ + bβSγαρ + cγSαβρ)/Sαβγ. (b) [aαSβγρ + (aβ + bα)Sγαρ + (cγSαβρ]/Sαβγ. (c) [aαSβγρ + (aβ + bα)Sγαρ + (aγ + cβ)Sαβρ]/Sαβγ. (d) V ρ, qρq −1 . 17. Show that the latent roots of eρ − f V αρβ (f > 0, T α = T β = 1) are e + f , e + f Sαβ, e − f , corresponding to latent directions α + β, V αβ, α − β; and that this is therefore a general form for self-conjugate nonions. Determine the latent directions and roots in the limiting case when α = β, or −β or f = 0. 18. Show that the nonion of Ex. 16 takes the form bρ − f (αSβρ + βSαρ), where b is the mean latent root. 19. Substitute the nonion of Ex. 18 for φ in Ex. 6 and show that the quadric surface is cut in circles by planes perpendicular to α or β. When is the surface one of revolution? 20. If the conjugate of a nonion is its reciprocal, and the modulus is positive, then the nonion is a rotation; and conversely every rotation satisfies this condition. [If R, R−1 are conjugate nonions, then ρ2 = SρR−1 Rρ = SRρRρ = (Rρ)2 ; i.e., T Rρ/ρ = 1. Also Sρσ = SρR−1 Rσ = SRρRσ and therefore the angle between ρ, σ = ∠ between Rρ, Rσ. Therefore R strains a sphere with centre O into another sphere with centre O in which the angles between corresponding radii are equal and their order in space is the same, since mod R is positive. Hence the strain is a rotation.] (a) Show that (RφR−1 )n = Rφn R−1 . 21. Show that φ0 φ is self-conjugate, and that its latent roots are positive, and that therefore there are four real values of ψ that satisfy ψ 2 = φ0 φ, mod ψ = mod φ. [Let φ0 φi = ai; then a = −Siφ0 φi = (T φi)2 .] √ 22. If φ = Rψ, where ψ is the self-conjugate strain φ0 φ, then R is a rotation. So √ 0 −1 φ = χR, where χ = RψR = φφ . 23. Show that φ0 · V φβφγ = V βγ · mod φ. [56 j.] 24. Show that the strain φρ = ρ−aαSβρ, where α, β, are perpendicular unit vectors, consists of a shearing of all planes perpendicular to β, the amount and direction of sliding of each plane being aα per unit distance of the plane from O. 25. Determine ψ and R of Ex. 22 for the strain of Ex. 24, and find the latent directions and roots of ψ.
Chapter 5
PROJECT GUTENBERG ”SMALL PRINT”
End of The Project Gutenberg EBook of A Primer of Quaternions, by Arthur S. Hathaway *** END OF THE PROJECT GUTENBERG EBOOK A PRIMER OF QUATERNIONS *** This file should be named pqtrn10p.pdf or pqtrn10p.zip Corrected EDITIONS of our eBooks get a new NUMBER, pqtrn11p.pdf VERSIONS based on separate sources get new LETTER, tvrnt10ap.pdf E-text prepared by Cornell University, Joshua Hutchinson, John Hagerson, and the Online Distributed Proofreading Team. Project Gutenberg eBooks are often created from several printed editions, all of which are confirmed as Public Domain in the US unless a copyright notice is included. Thus, we usually do not keep eBooks in compliance with any particular paper edition. We are now trying to release all our eBooks one year in advance of the official release dates, leaving time for better editing. Please be encouraged to tell us about any error or corrections, even years after the official publication date. Please note neither this listing nor its contents are final til midnight of the last day of the month of any such announcement. The official release date of all Project Gutenberg eBooks is at Midnight, Central Time, of the last day of the stated month. A preliminary version may often be posted for suggestion, comment and editing by those who wish to do so.
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT” Most people start at our Web sites at: http://gutenberg.net or http://promo.net/pg These Web sites include award-winning information about Project Gutenberg, including how to donate, how to help produce our new eBooks, and how to subscribe to our email newsletter (free!).
Those of you who want to download any eBook before announcement can get to them as follows, and just download by date. This is also a good way to get them instantly upon announcement, as the indexes our cataloguers produce obviously take a while after an announcement goes out in the Project Gutenberg Newsletter. http://www.ibiblio.org/gutenberg/etext03 or ftp://ftp.ibiblio.org/pub/docs/books/gutenberg/etext03 Or /etext02, 01, 00, 99, 98, 97, 96, 95, 94, 93, 92, 92, 91 or 90 Just search by the first five letters of the filename you want, as it appears in our Newsletters.
Information about Project Gutenberg (one page) We produce about two million dollars for each hour we work. The time it takes us, a rather conservative estimate, is fifty hours to get any eBook selected, entered, proofread, edited, copyright searched and analyzed, the copyright letters written, etc. Our projected audience is one hundred million readers. If the value per text is nominally estimated at one dollar then we produce $2 million dollars per hour in 2002 as we release over 100 new text files per month: 1240 more eBooks in 2001 for a total of 4000+ We are already on our way to trying for 2000 more eBooks in 2002 If they reach just 1-2% of the world’s population then the total will reach over half a trillion eBooks given away by year’s end. The Goal of Project Gutenberg is to Give Away 1 Trillion eBooks! This is ten thousand titles each to one hundred million readers, which is only about 4% of the present number of computer users. Here is the briefest record of our progress (* means estimated): eBooks Year Month 1 10 100 1000
1971 1991 1994 1997
July January January August
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT” 1500 2000 2500 3000 4000 6000 9000 10000
1998 1999 2000 2001 2001 2002 2003 2004
October December December November October/November December* November* January*
The Project Gutenberg Literary Archive Foundation has been created to secure a future for Project Gutenberg into the next millennium. We need your donations more than ever! As of February, 2002, contributions are being solicited from people and organizations in: Alabama, Alaska, Arkansas, Connecticut, Delaware, District of Columbia, Florida, Georgia, Hawaii, Illinois, Indiana, Iowa, Kansas, Kentucky, Louisiana, Maine, Massachusetts, Michigan, Mississippi, Missouri, Montana, Nebraska, Nevada, New Hampshire, New Jersey, New Mexico, New York, North Carolina, Ohio, Oklahoma, Oregon, Pennsylvania, Rhode Island, South Carolina, South Dakota, Tennessee, Texas, Utah, Vermont, Virginia, Washington, West Virginia, Wisconsin, and Wyoming. We have filed in all 50 states now, but these are the only ones that have responded. As the requirements for other states are met, additions to this list will be made and fund raising will begin in the additional states. Please feel free to ask to check the status of your state. In answer to various questions we have received on this: We are constantly working on finishing the paperwork to legally request donations in all 50 states. If your state is not listed and you would like to know if we have added it since the list you have, just ask. While we cannot solicit donations from people in states where we are not yet registered, we know of no prohibition against accepting donations from donors in these states who approach us with an offer to donate. International donations are accepted, but we don’t know ANYTHING about how to make them tax-deductible, or even if they CAN be made deductible, and don’t have the staff to handle it even if there are ways. Donations by check or money order may be sent to:
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT”
Project Gutenberg Literary Archive Foundation PMB 113 1739 University Ave. Oxford, MS 38655-4109 Contact us if you want to arrange for a wire transfer or payment method other than by check or money order. The Project Gutenberg Literary Archive Foundation has been approved by the US Internal Revenue Service as a 501(c)(3) organization with EIN [Employee Identification Number] 64-622154. Donations are tax-deductible to the maximum extent permitted by law. As fund-raising requirements for other states are met, additions to this list will be made and fund-raising will begin in the additional states. We need your donations more than ever! You can get up to date donation information online at: http://www.gutenberg.net/donation.html
*** If you can’t reach Project Gutenberg, you can always email directly to: Michael S. Hart
Prof. Hart will answer or forward your message. We would prefer to send you information by email.
**The Legal Small Print**
(Three Pages) ***START**THE SMALL PRINT!**FOR PUBLIC DOMAIN EBOOKS**START*** Why is this "Small Print!" statement here? You know: lawyers. They tell us you might sue us if there is something wrong with your copy of this eBook, even if you got it for free from someone other than us, and even if what’s wrong is not our fault. So, among other things, this "Small Print!" statement disclaims most of our liability to you. It also tells you how you may distribute copies of this eBook if you want to. *BEFORE!* YOU USE OR READ THIS EBOOK
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT” By using or reading any part of this PROJECT GUTENBERG-tm eBook, you indicate that you understand, agree to and accept this "Small Print!" statement. If you do not, you can receive a refund of the money (if any) you paid for this eBook by sending a request within 30 days of receiving it to the person you got it from. If you received this eBook on a physical medium (such as a disk), you must return it with your request. ABOUT PROJECT GUTENBERG-TM EBOOKS This PROJECT GUTENBERG-tm eBook, like most PROJECT GUTENBERG-tm eBooks, is a "public domain" work distributed by Professor Michael S. Hart through the Project Gutenberg Association (the "Project"). Among other things, this means that no one owns a United States copyright on or for this work, so the Project (and you!) can copy and distribute it in the United States without permission and without paying copyright royalties. Special rules, set forth below, apply if you wish to copy and distribute this eBook under the "PROJECT GUTENBERG" trademark. Please do not use the "PROJECT GUTENBERG" trademark to market any commercial products without permission. To create these eBooks, the Project expends considerable efforts to identify, transcribe and proofread public domain works. Despite these efforts, the Project’s eBooks and any medium they may be on may contain "Defects". Among other things, Defects may take the form of incomplete, inaccurate or corrupt data, transcription errors, a copyright or other intellectual property infringement, a defective or damaged disk or other eBook medium, a computer virus, or computer codes that damage or cannot be read by your equipment. LIMITED WARRANTY; DISCLAIMER OF DAMAGES But for the "Right of Replacement or Refund" described below, [1] Michael Hart and the Foundation (and any other party you may receive this eBook from as a PROJECT GUTENBERG-tm eBook) disclaims all liability to you for damages, costs and expenses, including legal fees, and [2] YOU HAVE NO REMEDIES FOR NEGLIGENCE OR UNDER STRICT LIABILITY, OR FOR BREACH OF WARRANTY OR CONTRACT, INCLUDING BUT NOT LIMITED TO INDIRECT, CONSEQUENTIAL, PUNITIVE OR INCIDENTAL DAMAGES, EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH DAMAGES. If you discover a Defect in this eBook within 90 days of receiving it, you can receive a refund of the money (if any) you paid for it by sending an explanatory note within that time to the person you received it from. If you received it on a physical medium, you must return it with your note, and such person may choose to alternatively give you a replacement copy. If you received it electronically, such person may
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT” choose to alternatively give you a second opportunity to receive it electronically. THIS EBOOK IS OTHERWISE PROVIDED TO YOU "AS-IS". NO OTHER WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, ARE MADE TO YOU AS TO THE EBOOK OR ANY MEDIUM IT MAY BE ON, INCLUDING BUT NOT LIMITED TO WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Some states do not allow disclaimers of implied warranties or the exclusion or limitation of consequential damages, so the above disclaimers and exclusions may not apply to you, and you may have other legal rights. INDEMNITY You will indemnify and hold Michael Hart, the Foundation, and its trustees and agents, and any volunteers associated with the production and distribution of Project Gutenberg-tm texts harmless, from all liability, cost and expense, including legal fees, that arise directly or indirectly from any of the following that you do or cause: [1] distribution of this eBook, [2] alteration, modification, or addition to the eBook, or [3] any Defect. DISTRIBUTION UNDER "PROJECT GUTENBERG-tm" You may distribute copies of this eBook electronically, or by disk, book or any other medium if you either delete this "Small Print!" and all other references to Project Gutenberg, or: [1]
Only give exact copies of it. Among other things, this requires that you do not remove, alter or modify the eBook or this "small print!" statement. You may however, if you wish, distribute this eBook in machine readable binary, compressed, mark-up, or proprietary form, including any form resulting from conversion by word processing or hypertext software, but only so long as *EITHER*: [*]
The eBook, when displayed, is clearly readable, and does *not* contain characters other than those intended by the author of the work, although tilde (~), asterisk (*) and underline (_) characters may be used to convey punctuation intended by the author, and additional characters may be used to indicate hypertext links; OR
[*]
The eBook may be readily converted by the reader at no expense into plain ASCII, EBCDIC or equivalent form by the program that displays the eBook (as is
CHAPTER 5. PROJECT GUTENBERG ”SMALL PRINT” the case, for instance, with most word processors); OR [*]
You provide, or agree to also provide on request at no additional cost, fee or expense, a copy of the eBook in its original plain ASCII form (or in EBCDIC or other equivalent proprietary form).
[2]
Honor the eBook refund and replacement provisions of this "Small Print!" statement.
[3]
Pay a trademark license fee to the Foundation of 20% of the gross profits you derive calculated using the method you already use to calculate your applicable taxes. If you don’t derive profits, no royalty is due. Royalties are payable to "Project Gutenberg Literary Archive Foundation" the 60 days following each date you prepare (or were legally required to prepare) your annual (or equivalent periodic) tax return. Please contact us beforehand to let us know your plans and to work out the details.
WHAT IF YOU *WANT* TO SEND MONEY EVEN IF YOU DON’T HAVE TO? Project Gutenberg is dedicated to increasing the number of public domain and licensed works that can be freely distributed in machine readable form. The Project gratefully accepts contributions of money, time, public domain materials, or royalty free copyright licenses. Money should be paid to the: "Project Gutenberg Literary Archive Foundation." If you are interested in contributing scanning equipment or software or other items, please contact Michael Hart at: [email protected] [Portions of this eBook’s header and trailer may be reprinted only when distributed free of all fees. Copyright (C) 2001, 2002 by Michael S. Hart. Project Gutenberg is a TradeMark and may not be used in any sales of Project Gutenberg eBooks or other materials be they hardware or software or any other related product without express permission.] *END THE SMALL PRINT! FOR PUBLIC DOMAIN EBOOKS*Ver.02/11/02*END*
